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.

Question

Quizplus · Accepted Answer

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

Solve the Problem (x−3)2+(y+5)2=36( x - 3 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 36(x−3)2+(y+5)2=36

Solve the Problem $( x - 3 ) ^ { 2 } + ( y + 5 ) ^ { 2 } = 36$