Circles

Geometry Level 4

Two circles $S_1$ and $S_2$ pass through the points $(0,a) , (0,-a)$ . The line $y=mx+c$ is tangent to the two circles . If $S_{1}$ and $S_{2}$ are orthogonal , then

c^{2}=a^{2}(4+m^{2})

c^{2}=a^{2}(3+m^{2})

c^{2}=a^{2}(1+m^{2})

c^{2}=a^{2}(2+m^{2})

1 solution

Garrett Clarke
Jul 7, 2015

The equation should work for any $m$ , correct? We'll make two circles of equal radius $r$ , where both circles go through the points $(0,a$ ) and $(0,-a)$ . Please refer to the image below when reading the solution.

In order for two circles to be orthogonal, then we must have the equation $r_1^2+r_2^2=d^2$ , where $d$ is the distance between the centers. Since $r_1=r_2=r$ in our setup, we have $2r^2=d^2$ . This makes our big triangle a 45-45-90 triangle.

We can see from our image that $a$ is an altitude, splitting the big triangle into two more 45-45-90 triangles. Since $a$ is the leg and $r$ is our hypotenuse in this case, we have the equation $r^2=2a^2$

Finally, clearly the equation of our line is $y=c$ , because the slope of our line is $0$ . Also, our picture shows that $c=r$ as the line is tangent to each circle and parallel to the $x$ -axis. Plugging in $c$ into our equation above:

$c^2=2a^2 \Longrightarrow c^2=a^2(2+m^2)$ where $m=0$ .

Therefore our answer must be $\boxed{c^2=a^2(2+m^2)}$ .

This is true for a special case, but not be true in general !! Assumption is that it is true for all cases but I think this type of proof is not taken as a good proof.

Niranjan Khanderia - 3 years, 9 months ago

Circles

1 solution

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