Question

The point (6,-10) is on a circle with center (2,-6). Write the standard equation of the circle.

Answer 1

circle equation with center (a, b) and radius r:
\[(x - a)^{2} + (y - b)^{2} = r^{2}\]
center (2, -6):
\[(x - 2)^{2} + (y +6)^{2} = r^{2}\]
passing through (6, -10):
\[(6 - 2)^{2} + (-10 +6)^{2} = r^{2}\]
\[4^{2} + (-4)^{2} = r^{2}\]
\[16 + 16 = r^{2}\]
\[32 = r^{2}\]
so, the equation of the circle is:
\[(x - 2)^{2} + (y +6)^{2} = 32\]
\[x^{2} - 2x - 2x + 4 + y^{2} + 6y + 6y + 36 = 32\]
\[x^{2} + y^{2} - 4x + 12y + 8 = 0\]