Question

A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

a. Find the center of the circle.
b. Find the radius. If your answer is not an integer, express it in radical form.
c. Write an equation for the circle.

Answer 1

a. (-3,2)

b. sqrt(65)

c. (x + 3)² + (y - 2)² = 65

Explanation:

Centre:

midpoint of the diameter

(h,k) = (-10+4)/2, (-2+6)/2

(h,k) = (-3,2)

Length of the diameter:

sqrt[(-2-6)² + (-10-4)²]

sqrt(260)

2sqrt(65)

Radius: ½ × diameter

Radius = sqrt(65)

Equation:

(x - h)² + (y - k)² = r²

(x - -3)² + (y - 2)² = (sqrt(65))²

(x + 3)² + (y - 2)² = 65

Answer 2

a. (-3, 2).

b. √65

c. (x + 3)^2 + (y - 2)^2 = 65

Explanation:

a. The center is the midpoint of the diameter PQ.

= (-10+4)/2, (-2+6)/2

= (-3, 2).

b. The radius is the distance from the center to a point on the circle.

Take the point (4, 6):

The radius = √((-3-4)^2 + (2-6)^2)

= √65.

c. The equation of the circle is:

Using the standard form

(x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center and r = the radius:

it is (x - (-3)^2 + (y - 2) = 65

= (x + 3)^2 + (y - 2)^2 = 65.