Find the equation of the circle: y-intercepts 4 and –8, contain (–12, –8)

  (x +6)^2 +(y +2)^2 = 72

Step-by-step explanation:

The given points are vertices of a right triangle. The circle circumscribing that triangle (through all 3 vertices) will have its center at the midpoint of the hypotenuse:

  ((0, 4) +(-12, -8))/2 = (-6, -2)

The equation of a circle with center (h, k) through point (a, b) is ...

  (x -h)^2 +(y -k)^2 = (a -h)^2 +(b -k)^2

For center (-6, -2) and point (0, 4), the equation is ...

  (x +6)^2 +(y +2)^2 = (0+6)^2 +(4 +2)^2

  (x +6)^2 +(y +2)^2 = 72

mujhko hui na

set the two equations for y equal to each other. solve for x. this will be the x-coordinate for the point of intersection. use this x-coordinate and plug it into either of the original equations for the lines and solve for y.