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I need to find a point on the $y-axis$ so the tangents from that point to circles:

$(x-6)^2+(y-3)^2=16$, $(x-4)^2+(y-6)^2=5$ are equal in length.

I tried to use $(x-a)(x_1-a)+(y-b)(y_1-b)$ but it was quite messy. i also tried to use $(-ma+b-n)^2=R^2(m^2+1)$ but couldn't see the end of it. i tried to use distances some how but it was too complex in the calculation part.

Thanks for the help.

Gerry Myerson
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debii
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1 Answers1

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It is useful to make a sketch of the two circles. Let the point we are looking for be $P=(0,b)$.

By the Pythagorean Theorem, the square of the distance from $P$ to the points of tangency to the first circle is given by $$\left((0-3)^2+(b-6)^2\right) -16.$$ You can write a similar expression for the square of the distance from $P$ to the points of tangency with the second circle.

Equate the two squares of distance, and solve for $b$.

André Nicolas
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