Question

The equation of a circle is (x + 6)2 + (y - 4)2 = 16. The point (-6, 8) is on the circle.

What is the equation of the line that is tangent to the circle at (-6, 8)?

Answer 1

Know calculus?

Answer 2

noo

Answer 3

I think no need for calculus.....can you find the center of circle?

Answer 4

no

Answer 5

Ok, you are right, I see how you are going to solve.

Answer 6

(x-a)^2+(y-b)^2 =r^2 is the standard equation of circle with center at (a,b) and radius r
Now can you figure out value of (a,b)

Answer 7

@heygurl123 dere?

Answer 8

huh?

Answer 9

can you find the center of circle now?

Answer 10

still no

Answer 11

(x-a)^2+(y-b)^2 =r^2 is the standard equation
(x + 6)^2 + (y - 4)^2 = 16 in the question can be written as (x -(- 6))^2 + (y -( 4))^2 = 4^2

Answer 12

I think now you can find value of (a,b)

Answer 13

@ankit042 made it pretty clear
for every circle of form
\[(x-a)^{2} +(y-b)^{2} = r^{2}\]
center is point (a,b)
radius is r
you are given
\[(x+6)^{2} + (y-4)^{2} = 16\]
x+6 = x-a
--> a = -6

y-4 = y-b
--> b=4

r^2 = 16
r = 4