Answer:
Suppose f(x,y)=x2+y2
Let’s look at the partial derivatives of this function:
∂f∂x=2x
∂f∂y=2y
So apparently, at each (x,y) coordinate pair (black dots below), the gradient of f(x,y) is pointing in the same direction as the position vector itself (it is only twice as large). Knowing that an isocline f(x,y)=C (a curve on which f(x,y) is constant) must be perpendicular to the gradient of f(x,y) . This yields the following image:
This image depicts for two position vectors (black dots), the gradient (blue vectors), and the perpendicular direction in which f(x,y) is constant (red line fragments).
Doing this for many positions we see that this creates circles:
Therefore setting f(x,y)=C , yields a circle in the 2D plane, with radius r=C−−√ .
EDIT: again, thanks to Gilles Castel for the nice graphics!
A conic section (or simply conic) is a curve obtained as the intersection of the surface of a cone with a plane. The three types of conic section are thehyperbola, the parabola, and the ellipse. The circle is a special case of the ellipse, and is of sufficient interest in its own right that it was sometimes called a fourth type of conic section.
In the Cartesian coordinate system, the graph of a quadratic equation in two variables is always a conic section and all conic sections arise in this way. The most general equation is of the form
The conic sections described by this equation can be class
Consider a triangle with hypotenuse r and angle θ . As θ goes from 0 to 2π it traces out a circle.
The x and y coordinates are give by the basic trig equations:
x=rsin(θ)
y=rcos(θ)
Now if we compute x2+y2 we get
x2+y2
=r2sin2(θ)+r2cos2(θ)
=r2(sin2(θ)+cos2(θ))
=r2(1)
Hence x2+y2=r2 .
Please help i need to find X.
Answer:
(11√42)/2
Step-by-step explanation:
Thanks to the given information we can say that the smaller leg of the left triangle is
(11√7)/2
while the other leg is
11/2 √7 * √3 = 11/2 √21
the triangle on the right has two congruent legs. For the Pythagorean theorem
x^2 = 2 (11/2 √21)^2
x^2 = 2 * 121/4 * 21
x^2 = 2541 / 2
x = √11^2 * 3 * 7/√2
x = 11√21/√2
x = (11√42)/2
Nadia is mountain climbing. She started at an altitude of 19.26 feet below sea level and then changed her altitude by climbing a total of 5,437.8 feet up from her initial position. What was Nadia’s altitude at the end of her climb?
Answer:
5418.54 ft
Step-by-step explanation:
So sea level is 0, okay? So Nadia (her name is more than 3 letters and I'm lazy so from now on she'll be reffered to as "N") is at -19. 26 ft. N goes up 5,437.8 ft, so we add this value on.
-19. 26+ 5437.8= 5418.54
Now just add on the units!
Hope this helps!
Answer:
Answer:
5418.54 ft
Step-by-step explanation:
Answer:
5418.54 ft
Step-by-step explanation:
So sea level is 0, okay? So Nadia (her name is more than 3 letters and I'm lazy so from now on she'll be reffered to as "N") is at -19. 26 ft. N goes up 5,437.8 ft, so we add this value on.
-19. 26+ 5437.8= 5418.54
Find the distance between each pair of points. Round to the nearest tenth if necessary.
(4,2) and (-6, -6)
Answer:
Radical (20)
Step-by-step explanation:
Radical ( (4-6)² + (2-6)²)) =radical ( 4+16) = radical (20)
The line y = 2x + 6 cuts the x-axis at A and the y-axis at B. Find
(a) the length of AB,
(b) the shortest distance of O to AB, where O is the origin (0,0)
Answer:
(a)
[tex]3 \sqrt{5} [/tex]
(b)
[tex] \frac{6}{ \sqrt{5} } [/tex]
Step-by-step explanation:
A(-3,0)
B(0,6)
[tex]d = \sqrt{{( - 3 - 0)}^{2} + {(0 - 6)}^{2} } = \sqrt{9 + 36} = 3 \sqrt{5} [/tex]
[tex]d = \frac{ax0 + by0 + c}{ \sqrt{ {a}^{2} + {b}^{2} } } [/tex]
2x-y+6=0
a=2, b=-1, c=6
x0=0, y0=0
[tex]d = \frac{6}{ \sqrt{4 + 1} } = \frac{6}{ \sqrt{5} } [/tex]
Find the missing side. Round to the nearest tenth. (ITS DUE IN THE MORNING PLEASE HELP)
24.
A) 14.2
C) 13.8
B) 9.2
D) 15.7.
25.
A) 37.6
B)30.8
C) 45.1
D)5.5
Answer:
24. A)14.2
25. D)5.5
Step-by-step explanation:
I hope it help for you keep safe
Oof, someone please help asap! I don't recall ever seeing this type of question before!
Answer:
90
Step-by-step explanation:
You are looking for the highest common factor for 270 and 360.
Factor the 2 numbers
270 = 2 * 5 * 3 * 3 * 3
360 = 2 * 2 * 2 * 3 * 3 * 5
Each of the numbers has a 5
Each of the numbers has two threes
Each of the numbers has one 2
So the answer is 2 * 3*3 * 5
Which of the following is the function for the graph shown?
Answer:
D
Step-by-step explanation:
The zeros from the graph, where it crosses the x- axis are
x = - 2 and x = 3 , then the corresponding factors are
(x + 2) and (x - 3) , then
y = a(x + 2)(x - 3) ( where a is a multiplier )
To find a substitute any point on the graph into the equation
Using (0, - 6 )
- 6 = a(0 + 2)(0 - 3) = a(2)(- 3) = - 6a ( divide both sides by - 6 )
1 = a
y = (x + 2)(x - 3) ← expand using FOIL
y = x² - x - 6 → D
Find all possible values of α+
β+γ when tanα+tanβ+tanγ = tanαtanβtanγ (-π/2<α<π/2 , -π/2<β<π/2 , -π/2<γ<π/2)
Show your work too. Thank you!
Answer:
[tex]\rm\displaystyle 0,\pm\pi [/tex]
Step-by-step explanation:
please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation
===========================
we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first
cancel tanγ from both sides which yields:
[tex] \rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma ) [/tex]
factor out tanγ:
[tex]\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)[/tex]
divide both sides by tanαtanβ-1 and that yields:
[tex]\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}[/tex]
multiply both numerator and denominator by-1 which yields:
[tex]\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)[/tex]
recall angle sum indentity of tan:
[tex]\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta ) [/tex]
let α+β be t and transform:
[tex]\rm\displaystyle \tan( \gamma ) = - \tan( t) [/tex]
remember that tan(t)=tan(t±kπ) so
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi ) [/tex]
therefore when k is 1 we obtain:
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi ) [/tex]
remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus
[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi ) [/tex]
recall that if we have common trigonometric function in both sides then the angle must equal which yields:
[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm \pi [/tex]
isolate -α-β to left hand side and change its sign:
[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }[/tex]
when is 0:
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 ) [/tex]
likewise by Opposite Angle Identity we obtain:
[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 ) [/tex]
recall that if we have common trigonometric function in both sides then the angle must equal therefore:
[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm 0 [/tex]
isolate -α-β to left hand side and change its sign:
[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }[/tex]
and we're done!
Answer:
-π, 0, and π
Step-by-step explanation:
You can solve for tan y :
tan y (tan a + tan B - 1) = tan a + tan y
Assuming tan a + tan B ≠ 1, we obtain
[tex]tan/y/=-\frac{tan/a/+tan/B/}{1-tan/a/tan/B/} =-tan(a+B)[/tex]
which implies that
y = -a - B + kπ
for some integer k. Thus
a + B + y = kπ
With the stated limitations, we can only have k = 0, k = 1 or k = -1. All cases are possible: we get k = 0 for a = B = y = 0; we get k = 1 when a, B, y are the angles of an acute triangle; and k = - 1 by taking the negatives of the previous cases.
It remains to analyze the case when "tan "a" tan B = 1, which is the same as saying that tan B = cot a = tan(π/2 - a), so
[tex]B=\frac{\pi }{2} - a + k\pi[/tex]
but with the given limitation we must have k = 0, because 0 < π/2 - a < π.
On the other hand we also need "tan "a" + tan B = 0, so B = - a + kπ, but again
k = 0, so we obtain
[tex]\frac{\pi }{2} - a=-a[/tex]
a contradiction.
find the slope of the line passing through the points (-2,5) and (3/2,2)
Answer:
slope = - [tex]\frac{6}{7}[/tex]
Step-by-step explanation:
Calculate the slope m using the slope formula
m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]
with (x₁, y₁ ) = (- 2, 5) and (x₂, y₂ ) = ([tex]\frac{3}{2}[/tex], 2)
m = [tex]\frac{2-5}{\frac{3}{2}-(-2) }[/tex]
= [tex]\frac{-3}{\frac{3}{2}+2 }[/tex]
= [tex]\frac{-3}{\frac{7}{2} }[/tex]
= - 3 × [tex]\frac{2}{7}[/tex]
= - [tex]\frac{6}{7}[/tex]
Find x in the kite below
Answer:
5
Step-by-step explanation:
The given is 3-4-5 special triangle so the missing side represented with x is 5
For each triangle shown below, determine whether you would use the Law of Sines or Law of Cosines to find angle x, and explain how you know which Law to use. Then find angle x to the nearest tenth.
NOTE: The perimeter of ABC-31
Given:
The figure of a triangle.
The perimeter of the triangle ABC is 31.
To find:
The value of x in the given triangle.
Solution:
Three sides of the triangle ABC are AB, BC, AC are their measures are [tex]3b-4,2b+1,b+10[/tex] respectively.
The perimeter of the triangle ABC is 31.
[tex]AB+BC+AC=31[/tex]
[tex](3b-4)+(2b+1)+(b+10)=31[/tex]
[tex]6b+7=31[/tex]
Subtract 7 from both sides.
[tex]6b=31-7[/tex]
[tex]6b=24[/tex]
[tex]b=\dfrac{24}{6}[/tex]
[tex]b=4[/tex]
Now, the measures of the sides are:
[tex]AB=3b-4[/tex]
[tex]AB=3(4)-4[/tex]
[tex]AB=12-4[/tex]
[tex]AB=8[/tex]
[tex]BC=2b+1[/tex]
[tex]BC=2(4)+1[/tex]
[tex]BC=8+1[/tex]
[tex]BC=9[/tex]
And,
[tex]AC=b+10[/tex]
[tex]AC=4+10[/tex]
[tex]AC=14[/tex]
Using the law of cosines, we get
[tex]\cos A=\dfrac{b^2+c^2-a^2}{2bc}[/tex]
[tex]\cos A=\dfrac{(AC)^2+(AB)^2-(BC)^2}{2(AC)(AB)}[/tex]
[tex]\cos A=\dfrac{(14)^2+(8)^2-(9)^2}{2(14)(8)}[/tex]
[tex]\cos A=\dfrac{179}{224}[/tex]
Using calculator, we get
[tex]\cos A=0.7991[/tex]
[tex]A=\cos ^{-1}(0.7991)[/tex]
[tex]x=36.9558^\circ[/tex]
[tex]x\approx 37.0^\circ[/tex]
Therefore, the value of x is 37.0 degrees.
4 Two number cubes are rolled. The faces have the numbers 1-6 on them. The number that shows on top is recorded. What is the probability that
the same number shows on both number cubes?
OF
OS
What kind of statement does the shorthand below represent?
Answer:
it's a transitive statement.
Similar to "If a is equal to b and b is equal to c, then a is equal to c."
a = b and b = c, then a = c
Step-by-step explanation:
Evaluate the integral.
Answer:
D
Step-by-step explanation:
[tex]\int\limits^4_1 {\frac{7^{lnx}}{x} } \, dx \\put~ln~x=y\\diff.\\\frac{1}{x} dx=dy\\when~x=1,y=ln1=0\\when x=4,y=ln~4\\\int\limits^{ln4}_0 {7^y} \, dy\\=\frac{7^y}{ln7} |0 ~\rightarrow~ ln 4\\= \frac{7^{ln4}}{7}[/tex]
help or i will fail my acellus
Answer:
I think it's 155 cm
Step-by-step explanation:
=(5×5×3)+(5×2×4)
= 75+40
= 155 cm2
1. Paul uses a coordinate plane to design
his model town layout.
Paul moves the market 2 units left and 3
units down. He says the ordered pair for
the new location of the market is (0,6).
Explain Paul's mistake and write the
correct ordered pair for the new location of
the market.
PLZ ALSO INCLUDE WHAT HIS MISTAKE WAS!
ANSWER FOR Brainiest!!!
The graph of y=x^3+x^2-6x is shown....
hello,
" a turning point is defined as the point where a graph changes from either increasing to decreasing, or decreasing to increasing"
a)
[tex]y=x^3+x^2-6x\\\\y'=3x^2+2x-6=0\\x=\dfrac{-2-\sqrt{76} }{6} \approx{-1.786299647...}\\or\\x=\dfrac{-2+\sqrt{76} }{6} \approx{1.1196329...}\\[/tex]
b)
Zeros are -3,0,2.
Sol={-3,0,2}
The solution of the graph function y=x³+x²-6x are -3 , 0 and 2
What is graph?The link between lines and points is described by a graph, which is a mathematical description of a network. A graph is made up of certain points and the connecting lines. It doesn't matter how long the lines are or where the points are located. A node is the name for each element in a graph.
We have the function
y=x³+x²-6x
now, equating it to 0
x³+x²-6x = 0
x² + x - 6= 0
x² - 3x + 2x -6 =0
x(x -3) + 2(x -3)
x= 3 and -2
Now, ew can see from the that the equation is touching the x-axis at three points and it will represent three zeroes of the equation.
So, the solution of the graph are -3 , 0 and 2
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Seo-Yun organizó una fiesta. Comprar 50 recuerditos para regalar y les dio 3 recuerditos a cada uno de sus invitados conforme llegaban a la fiesta.
Escribe una fórmula explícita para la sucesión.
g(n)=
Answer: nose
Step-by-step explanation:
8,X,20 are in arithmetic progression,find the value of "x".
Answer:
x = 14
Step-by-step explanation:
Since the terns form an arithmetic progression then they have a common difference d , that is
a₂ - a₁ = a₃ - a₂
x - 8 = 20 - x ( add x to both sides )
2x - 8 = 20 ( add 8 to both sides )
2x = 28 ( divide both sides by 2 )
x = 14
Question 4 Multiple Choice Worth 4 points)
(01.02 LC)
What is the solution for the equation 6x - 8 = 4x?
Answer:
Algebra
Step-by-step explanation:
(+) it's the same thing btw6x -8 = 4x
(collect like terms) meaning the numbers with x go over the " = " sign
making it = -8 = 4x -6x
the signs change when it crosses over so it becomes that
-8 = -2x
-8 ÷ -2 = 4 (cause - ÷ by - is + )
???????????????????????????
Answer: its 20 I think
Answer:
x = 50
I hope this help the side note also help me a lot as well
Triangle ABC is congruent to LMN. Find the value of x. Please and thank you!
Warning: if you give an answer that is NOT related to the question at all then I will report you - FIND THE VALUE OF X
Answer:
x = 9
Step-by-step explanation:
The ratios of corresponding sides are equal, that is
[tex]\frac{BC}{MN}[/tex] = [tex]\frac{AB}{LM}[/tex] , substitute values
[tex]\frac{x}{15}[/tex] = [tex]\frac{6}{10}[/tex] ( cross- multiply )
10x = 90 ( divide both sides by 10 )
x = 9
which of the following are identities? check all that apply.
A. (sinx + cosx)^2= 1+sin2x
B. sin6x=2 sin3x cos3x
C. sin3x/sinxcosx = 4cosx - secx
D. sin3x-sinx/cos3x+cosx = tanx
Answer: (a), (b), (c), and (d)
Step-by-step explanation:
Check the options
[tex](a)\\\Rightarrow [\sin x+\cos x]^2=\sin ^2x+\cos ^2x+2\sin x\cos x\\\Rightarrow [\sin x+\cos x]^2=1+2\sin x\cos x\\\Rightarrow \Rightarrow [\sin x+\cos x]^2=1+\sin 2x[/tex]
[tex](b)\\\Rightarrow \sin (6x)=\sin 2(3x)\\\Rightarrow \sin 2(3x)=2\sin (3x)\cos (3x)[/tex]
[tex](c)\\\Rightarrow \dfrac{\sin 3x}{\sin x\cos x}=\dfrac{3\sin x-4\sin ^3x}{\sin x\cos x}\\\\\Rightarrow 3\sec x-4\sin ^2x\sec x\\\Rightarrow 3\sec x-4[1-\cos ^2x]\sec x\\\Rightarrow 3\sec x-4\sec x+4\cos x\\\Rightarrow 4\cos x-\sec x[/tex]
[tex](d)\\\Rightarrow \dfrac{\sin 3x-\sin x}{\cos 3x+\cos x}=\dfrac{2\cos [\frac{3x+x}{2}] \sin [\frac{3x-x}{2}]}{2\cos [\frac{3x+x}{2}]\cos [\frac{3x-x}{2}]}\\\\\Rightarrow \dfrac{2\cos 2x\sin x}{2\cos 2x\cos x}=\dfrac{\sin x}{\cos x}\\\\\Rightarrow \tan x[/tex]
Thus, all the identities are correct.
A. Not an identity
B. An identity
C. Not an identity
D. An identity
To check whether each expression is an identity, we need to verify if the equation holds true for all values of the variable x. If it is true for all values of x, then it is an identity. Let's check each option:
A. [tex]\((\sin x + \cos x)^2 = 1 + \sin 2x\)[/tex]
To check if this is an identity, let's expand the left-hand side (LHS):
[tex]\((\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x\)[/tex]
Now, we can use the trigonometric identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex] to simplify the LHS:
[tex]\(\sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x\)[/tex]
The simplified LHS is not equal to the right-hand side (RHS) 1 + sin 2x since it is missing the sin 2x term. Therefore, option A is not an identity.
B. [tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]
To check if this is an identity, we can use the double-angle identity for sine:[tex]\(\sin 2\theta = 2\sin \theta \cos \theta\)[/tex]
Let [tex]\(2\theta = 6x\)[/tex], which means [tex]\(\theta = 3x\):[/tex]
[tex]\(\sin 6x = 2 \sin 3x \cos 3x\)[/tex]
The equation holds true with the double-angle identity, so option B is an identity.
C. [tex]\(\frac{\sin 3x}{\sin x \cos x} = 4\cos x - \sec x\)[/tex]
To check if this is an identity, we can simplify the right-hand side (RHS) using trigonometric identities.
Recall that [tex]\(\sec x = \frac{1}{\cos x}\):[/tex]
[tex]\(4\cos x - \sec x = 4\cos x - \frac{1}{\cos x} = \frac{4\cos^2 x - 1}{\cos x}\)[/tex]
Now, using the double-angle identity for sine, [tex]\(\sin 2\theta = 2\sin \theta \cos \theta\),[/tex] let [tex]\(\theta = x\):[/tex]
[tex]\(\sin 2x = 2\sin x \cos x\)[/tex]
Multiply both sides by 2: [tex]\(2\sin x \cos x = \sin 2x\)[/tex]
Now, the left-hand side (LHS) becomes:
[tex]\(\frac{\sin 3x}{\sin x \cos x} = \frac{\sin 2x}{\sin x \cos x}\)[/tex]
Using the double-angle identity for sine again, let [tex]\(2\theta = 2x\):[/tex]
[tex]\(\frac{\sin 2x}{\sin x \cos x} = \frac{2\sin x \cos x}{\sin x \cos x} = 2\)[/tex]
So, the LHS is 2, which is not equal to the RHS [tex]\(\frac{4\cos^2 x - 1}{\cos x}\)[/tex]. Therefore, option C is not an identity.
D. [tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \tan x\)[/tex]
To check if this is an identity, we can use the sum-to-product trigonometric identities:
[tex]\(\sin A - \sin B = 2\sin \frac{A-B}{2} \cos \frac{A+B}{2}\)\(\cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2}\)[/tex]
Let A = 3x and B = x:
[tex]\(\sin 3x - \sin x = 2\sin x \cos 2x\)\(\cos 3x + \cos x = 2\cos 2x \cos x\)[/tex]
Now, we can rewrite the expression:
[tex]\(\frac{\sin 3x - \sin x}{\cos 3x + \cos x} = \frac{2\sin x \cos 2x}{2\cos 2x \cos x} = \frac{\sin x}{\cos x} = \tan x\)[/tex]
The equation holds true, so option D is an identity.
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Lines L and M are parallel.
Answer:
52°
Step-by-step explanation:
The angle 38° and m∠1 equal 90°, so 90-38=52.
Answer:
thats is an obtuse so if 2 is 38 then 1 should be 120 but u add those together u get 158 so it shold be 120 but if not then try looking explamles up
Step-by-step explanation:
help! urgent asap!!!!!!!!!!
Answer:
A
Step-by-step explanation:
This is because the answer you picked means congruent to, however these two triangles are not the same size
Help please guys if you don’t mind
Step-by-step explanation:
Step 2:
[tex] ({15x}^{2} - 9x) + (5x - 3)[/tex]
Stepc3:
[tex]3x(5x - 3)[/tex]
Come get your 11 points :)
Which expression is equivalent to 5g+9+8g+4?
[tex]\huge\textsf{Hey there!}[/tex]
[tex]\large\textsf{5g + 9 + 8g + 4}\\\\\huge\textsf{COMBINE the LIKE TERMS}\\\\\large\textsf{5g + 8g + 9 + 4}\\\\\large\textsf{5g + 8g = \bf 13g}\\\\\large\textsf{9 + 4 = \bf 13}\\\\\boxed{\huge\textsf{= 13g + 13}}\large\checkmark\\\\\\\\\\\\\boxed{\boxed{\huge\textsf{Answer: \bf 13g + 13}}}\huge\checkmark\\\\\\\\\\\\\\\\\large\textsf{Good luck on your assignment and enjoy your day!}\\\\\\\\\\\frak{Amphitrite1040:)}[/tex]
Which is the best estimate of 162% of 79?
I think it's answer is 79 just guess
HELP ASAP PLEASEEE !!!!
WILL GIVE BRAINLIEST
which equation represents the general form of a circle with a center at (-2 -3) and a diameter of 8 units ?
• x²+y²+4x+6y-51=0
•x²+y²-4x-6y-51=0
•x²+y²+4x+6y-3=0
•x²+y²-4x-6y-3=0
Answer:
it’s the 3rd one:)
Step-by-step explanation:
Answer:
In this situation, (h, k) = (-2, -3)Radius = 8/2 = 4 unitsSubstitute them into the circle equation: (x - h)² + (y - k)² = r²
(x - (-2))² + (y - (-3))² = 4²
(x + 2)² + (y + 3)² = 4²
Now expand the equation:
(x + 2)² + (y + 3)² = 4²
(x + 2)(x + 2) + (y + 3)(y + 3) = 16
(x² + 2x + 2x + 4) + (y² + 3y + 3y + 9) = 16
x² + 4x + 4 + y² + 6y + 9 = 16
x² + y² + 4x + 6y + 13 - 16 = 0
x² + y² + 4x + 6y - 3 = 0
