# If the abscissae and ordinates of two points P and Q are roots of the equations x2 + 2ax – b2 = 0 and x2 + 2px – q2 = 0 respectively, then write the equation of the circle with PQ as diameter.

Given that points, we need to find the equation of the circle whose ends of the diameter are P and Q.

It is also told that the abscissae of two points P and Q are the roots of x2 + 2ax - b2 = 0 and ordinates are the roots of x2 + 2px - q2 = 0.

Let us first find the roots of each quadratic equation.

For x2 + 2ax - b2 = 0

..... (1)

For x2 + 2px - q2 = 0

..... (2)

From (1) and (2) we get,

We know that the centre is the mid - point of the diameter.

Centre(C) =

C = (- a, - p)

We have a circle with centre (- a, - p) and passing through the point .

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

We know that the distance between the two points (x1,y1) and (x2,y2) is .

Let us assume r is the radius of the circle.

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

(x - p)2 + (y - q)2 = r2

Now we substitute the corresponding values in the equation:

x2 + 2ax + a2 + y2 + 2py + p2 = a2 + b2 + p2 + q2

x2 + y2 + 2ax + 2py - b2 - q2 = 0

The equation of the circle is x2 + y2 + 2ax + 2py - b2 - q2 = 0.

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