Draw up the equation of the circle passing through the point (5: 3) centered at the point of intersection

Draw up the equation of the circle passing through the point (5: 3) centered at the point of intersection of lines 5x-3y-13 = 0 and x + 4y + 2 = 0.

Find the point of intersection of lines 5x – 3y – 13 = 0 and x + 4y + 2 = 0.

Let us solve the system of equations for this:

5x – 3y – 13 = 0;

x + 4y + 2 = 0.

Substituting into the first equation the value x = -4y – 2 from the second equation, we get:

5 * (-4y – 2) – 3y – 13 = 0;

-20y – 10 – 3y – 13 = 0;

-23y – 23 = 0;

23y = -23;

y = -1.

Find x:

x = -4y – 2 = -4 * (-1) – 2 = 4 – 2 = 2.

Find the square of the radius of the circle, equal to the distance from the center of the circle (2; -1) to the point with coordinates (5; 3):

(2 – 5) ^ 2 + (-1 – 3) ^ 2) = 3 ^ 2 + 4 ^ 2 = 9 + 16 = 25.

We write the equation of a circle centered at a point (2; -1) of radius 5:

(x – 2) ^ 2 + (y + 1) ^ 2 = 25.

Answer: the equation of the circle is (x – 2) ^ 2 + (y + 1) ^ 2 = 25.