Question

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. 1) (x + 3)2 + (y + 5)2 = 36 2) (x + 3)2 + (y – 5)2 = 36 3) 6 - 3)2 + (y + 5)2 = 36 4) 6 - 3)2 + (x - 5)2 = 36

Answer 1

(x -3)² +(y -5)² = 36

Explanation:

The center is on the vertical line halfway between the given vertical lines, so is at x=3. It is also on the horizontal line through the point (-3, 5), so is at y=5. The center is 9-3=6 from either tangent line, so this is the radius.

For center (h, k) and radius r, the circle's equation is ...

(x -h)² +(y -k)² = r²

For (h, k) = (3, 5) and r=6, the equation is ...

(x -3)² +(y -5)² = 36 . . . . . . . . does not match any choice written here