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Problem 18

# The circles $$x^{2}+y^{2}+2 a x-c^{2}=0$$ and $$x^{2}+y^{2}+2 b x-c^{2}=0$$ intersect at $$A$$ and B. A line through $$A$$ meets one circle at $$P$$ and a parallel line through $$B$$ meets the other circle at Q. Show that the locus of the mid point of $$P Q$$ is a circle.

Expert verified
Hence, it has been proven that the locus of the midpoint of $$PQ$$ is a circle.
See the step by step solution

## Step 1: Find Intersection Points

Since circles intersect at $$A$$ and $$B$$, we can find the intersection points by setting the equations of the circles equal to each other. This gives $$2 a x - 2 b x = 0$$ implying $$x = 0$$. Substituting x = 0 in any of the circle equations will give $$y = c$$ and $$y = -c$$. Therefore the intersection points $$A$$ and $$B$$ are $$(0,c)$$ and $$(0,-c)$$ respectively.

## Step 2: Deduce Properties of Parallel Lines Intersecting Circles

By the nature of parallel lines, point $$P$$ on one circle and matched point $$Q$$ on the other not only form a similar structure but are mirror images with respect to the x-axis. Therefore, if $$P$$ has coordinates $$(x_{1}, y_{1})$$, then $$Q$$ will have coordinates $$(x_{1}, -y_{1})$$. Also, $$P$$ and $$Q$$ lie on the given circles, so they satisfy their equations.

## Step 3: Compute coordinates of Midpoint of $$PQ$$

The midpoint between two points $$(x_{1}, y_{1})$$ and $$(x_{2}, y_{2})$$ is given by $$\left( \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}\right)$$. Using this formula, the midpoint of $$PQ$$ will have coordinates $$\left(\frac{x_{1} + x_{1}}{2}, \frac{y_{1} - y_{1}}{2}\right)$$, simplifying this gives $$\left(x_{1}, 0\right)$$ which is a point on the x-axis.

## Step 4: Show that Locus is a Circle

A locus of points equidistant from a set point is a circle. In this case, the midpoints of $$PQ$$ are all on the x-axis ($$y = 0$$) and spread symmetrically around the origin. Their x-coordinates satisfy the equation of a circle centered at origin with radius $$\frac{a+b}{2}$$, which is $$x^{2} - (\frac{a+b}{2})^{2}=0$$. Hence the locus is a circle.

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