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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.

Short Answer

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Hence, it has been proven that the locus of the midpoint of \(PQ\) is a circle.
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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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