Equation of a circle problem with y-intercepts given

Question

given the $y$-intercepts $-18$ and $-8$, tangent to the $x-$axis. The answer is

$(x-12)^2+(y+13)^2=169$.

I just don't get it how to get the answer. The prof just gave the answer to it.

Answer 1

Hint:

The center $C$ is a point on the perpendicular bisector of the two $y$ intercepts that is the line $y=-13$, so $C=(\alpha,-13)$. Since the circle is tangent to the $x$ axis , its radius have to be $r=13$. This means that the distance from one $y$ intercept, e.g $(0,-8)$ is such that: $$ (0-\alpha)^2+(-8+13)^2=r^2= 169 $$

solve this for $\alpha$ and you have two circle.

Answer 2

Ok, this is what we do first: the line joining $(0,-18)$ and $(0,-8)$ forms a chord of the circle. Now, the perpendicular bisector of this chord is parallel to the x-axis, and has y-coordinate as $-13$. Hence because the perpendicular bisector of every chord passes through the center, the center has y-coordinate $13$.

Now, if the center has coordinates $(x,-13)$, then note that there is a right angled triangle, namely $(x,-13), (0,-13),(0,-18)$. Now, using the Pythagoras theorem, we get $x=\pm 12$.

Thus, we have two circles, one of center $(12,-13)$ and one of center $(-12,-13)$. Both have radius $13$, since they touch the x-axis.

Answer 3

Note: The problem seems to have two solutions: