Find Equation of a Circle by its Center and a point

How to find an Equation of a Circle by its Center and a point on it

QUESTION: What is the equation of the circle that has its center at

(-3, 4) and passes through the point (-3, 0)?

SOLUTION:

Standard equation of circle is:

(x – h)² + (y – k)² = r² ————-(A)

where (h, k) is the co-ordinate of center of a circle

and r is the radius of a circle.

Here,

(h, k) = (-3, 4)

r we have to find.

It is given that the circle passes through the point (-3, 0)

So, put

(x, y) = (-3, 0) (That is, x = -3 and y = 0)

and

(h, k) = (-3, 4) (That is, h = -3 and k = 4)

in equation (A) to find r.

(x – h)² + (y – k)² = r² —————–(A)

{-3 – (-3)}² + (0 – 4)² = r²

(-3 +3)² + (-4)² = r²

0 + 16 = r²

Taking square root on both sides

r = ±4

As radius is always positive so r=4.

Put values of h, k and r in equation (A)

(x – h)² + (y – k)² = r² ————(A)

{x – (-3)}² + (y – 4)² = 4²

(x + 3)² + (y – 4)² = 16

Thus, equation of the circle that has its center at (-3, 4) and passes through the point (-3, 0) is:

(x + 3)² + (y – 4)² = 16