Question

If (-4, 11) and (-6,5) are the endpoints of a diameter of a circle, what is the equation of the circle?

Answer 1

The equation of the circle is given as:

$(x+5)^2 + (y-8)^2 = 10$

Solution:

Given that,

(−4, 11) and (−6, 5) are the endpoints of a diameter of a circle

The standard form of the equation of a circle is:

$(x-a)^2 + (y-b)^2=r^2$

Where,

(a, b) are the co-ordinates of the centre and r is the radius

To find the centre:

Find the midpoint of two given points

$m = (\frac{x_1+x_2}{2} , \frac{y_1+y_2}{2})\\\\m = (\frac{-4-6}{ 2}, \frac{11+5}{2})\\\\m = (-5, 8)$

calculate the radius using the  distance formula

Distance between center and one end point = radius

(-5, 8) and (-4, 11)

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\\\d = \sqrt{(-4+5)^2 + (11-8)^2}\\\\d = \sqrt{1 + 9}\\\\d = \sqrt{10}$

The equation of the circle is given as:

$(x-a)^2 + (y-b)^2=r^2\\\\(x+5)^2 + (y-8)^2 = (\sqrt{10})^2\\\\(x+5)^2 + (y-8)^2 = 10$

Thus the equation of circle is found