Find the equation of the circle that is tangent to equation y=(-2) that has a center at (-6, 12)

Answer 1

The equation for the circle with center at (h,k) and radius r has the form (x-h)² + (y-k)² = r²

Since the circle is tangent to y=-2 and its center y coodinate is 12 the circle has a radius of 12-(-2) = 14.

Using the center’s coordinates we have then:

(x+6)² + (y-12)² = 14²

Answer 2

So, all circles have equations of the form

(x-h)²+(y-k)²=r²

where (h,k) is the center of the circle and r is the radius.

Therefore, you already know most of your equation.

(x+6)²+(y-12)²=r²

We know that the tangent to a circle is perpendicular to the radius at the point of tangency, so the radius from the center to the line y = -2 is perpendicular to the line y = -2. In other words, the radius is vertical. which means the x-coordinate of the point of tangency is the same as the x-coordinate of the center (drawing a picture here is very helpful). Anyway, the point of tangency is, therefore, (-6,-2). This means that (-6,-2) is on the circle, so you can substitute those values in for x and y and find the radius.

(-6+6)²+(-2-12)²=r²

196=r²

(x+6)²+(y-12)²=196