How do you find general form of circle tangent to the y-axis and has a center of (-6, 7)?

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Answer 1 · 2015-12-31T00:09:43

Use a little reasoning to find:

${(x + 6)}^{2} + {(y - 7)}^{2} = 6^{2}$ $(x+6)^2+(y-7)^2 = 6^2$

or if really fussy:

${(x - (- 6))}^{2} + {(y - 7)}^{2} = 6^{2}$ $(x-(-6))^2+(y-7)^2 = 6^2$

The general equation of a circle with centre $(h, k)$ $(h, k)$ and radius $r$ $r$ is:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2 = r^2$

In our case $(h, k) = (- 6, 7)$ $(h, k) = (-6, 7)$ and the radius $r$ $r$ must be $6$ $6$ for the circle to touch the $y$ $y$ axis.

graph{((x-(-6))^2+(y-7)^2-6^2)((x-(-6))^2+(y-7)^2-0.012)=0 [-24.75, 15.25, -3.52, 16.48]}