Solve the problem.
-Find the equation of a circle in standard form that is tangent to the line $x = - 3$ at $( - 3,5 )$ and also tangent to the line $x = 9$.
A) $( x - 3 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 36$
B) $( x + 3 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 36$
C) $( x + 3 ) ^ { 2 } + ( y - 5 ) ^ { 2 } = 36$
D) $( x - 3 ) ^ { 2 } + ( y - 5 ) ^ { 2 } = 36$

Question

Quizplus · Accepted Answer

The circle is tangent to the vertical lines $x = -3$ and $x = 9$, so its center is equidistant from these lines. The midpoint is at $x = 3$. The circle is tangent to $x = -3$ at $(-3, 5)$, so the radius is 6. The center is $(3, 5)$, and the equation is $(x - 3)^2 + (y - 5)^2 = 36$.

Solve the Problem x=−3x = - 3x=−3 At (−3,5)( - 3,5 )(−3,5)

Solve the Problem $x = - 3$ At $( - 3,5 )$