What is the equation of a circle that passes through (3,4) and is centered at (-3,6)?

Question

 i have no clue what i am doing please help me

Answer 1

equation for a circle is (x-h)^2+(y-k)^2=r^2

r= radius. The center is at (h,k), so h=-3, k=6:

(x-(-3))^2+(y-6)^2=r^2

(x+3)^2+(y-6)^2=r^2

plug in the x and y values for the known point on the circle and solve for r^2 to find the radius. x=3, y=4:

(3+3)^2+(4-6)^2=r^2

6^2+(-2)^2=r^2

36+4=r^2

40=r^2

 

 

so our equation is (x+3)^2+(y-6)^2=40

Answer 2

Hi Nicki,

 

The general equation for a circle centered at (x, y) = (h, k) is

 

(x - h)² + (y - k)² = r²

 

where r is the radius. For your case, the circle is centered about (h, k) = (-3, 6), so your equation becomes

 

(x + 3)² + (y - 6)² = r²

 

You also know that the circle goes through a certain point (x, y) = (3, 4). You can substitute for this particular solution and solve for r.

 

r² = (3 + 3)² + (4 - 6)² = 6² + (-2)² = 40

 

So, r = √40

 

Therefore, the equation for your circle is

 

(x + 3)² + (y - 6)² = 40