Given: 2 lines containing the diameter of a circle and a point lying on this circle; Find: the equation of this circle The lines $ y = \frac{4}{3}x - \frac{5}{3} $ and $ y = \frac{-4}{3}x - \frac{13}{3} $ each contain diameters of a circle. and the point $ (-5, 0) $ is also on that circle. 
Find the equation of this circle.
 A: Prerequisites

Given


*

*We have 2 lines with given equations (see below).


*

*$ y = \frac{4}{3}x - \frac{5}{3} $

*$ y = \frac{-4}{3}x - \frac{13}{3} $


*These lines each contain a diameter of the circle-in-question.

*We have a point with the given coordinate (see below).


*

*$ P = (-5, 0) $


*This point is on the circle in question.

Problem
Find the equation of the circle-in-question.
Here is our progress:
$ (x - h)^2 + (y - k)^2 = r^2 $

Solution

Step 1
Let us find the center of the circle-in-question.
Note that diameters of circles go through the center of a circle.
We can find the intersection of the two given lines.
Let us find the x-component of the center of the circle-in-question.
$ \frac{4}{3}x - \frac{5}{3} = \frac{-4}{3}x - \frac{13}{3} $
$ 4x - 5 = -4x - 13 $
$ 8x = -8 $
$ x = -1 $
Let us find the y-component of the center of the circle-in-question.
$
y 
= \frac{4}{3}x - \frac{5}{3} 
= \frac{4}{3}(-1) - \frac{5}{3} 
= \frac{-4}{3} - \frac{5}{3}
= \frac{-9}{3}
= -3
$
So...
The center of the circle-in-question is $ Q = (-1, -3) $.
Here is our progress:
$ (x - (-1))^2 + (y - (-3))^2 = r^2 $
$ (x + 1)^2 + (y + 3)^2 = r^2 $

Step 2
Let us find the radius of the circle-in-question.
We know that the radius of the circle-in-question is the distance between the center of the circle-in-question and the given point that lies on the circle-in-question.
Let us find the distance between such points.
$
l 
= \sqrt{((-1)-(-5))^2+((-3)-(0))^2}
= \sqrt{(4)^2+(-3)^2}
= \sqrt{16+9}
= \sqrt{25}
= 5
$
So...
The length of the radius is $ r = 5 $
Here is our progress:
$ (x + 1)^2 + (y + 3)^2 = (5)^2 $
$ (x + 1)^2 + (y + 3)^2 = 25 $

Answer

The equation of the circle-in-question is:
$$ (x + 1)^2 + (y + 3)^2 = 25 $$
