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
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