Question

If the circumference of the circle \[{{x}^{2}}+{{y}^{2}}-2x+8y-\lambda =0\] is bisected by the circle \[{{x}^{2}}+{{y}^{2}}+4x+22y+\lambda =0\], then \[\lambda \] equals.....

Answer 1

Hint:Find the centre of two circles and write the equation of common chord of both the circles. As the circumference of one circle is bisected by another circle, the centre of one circle must lie on the common chord. Use this concept to get the value of the parameter.

Complete step-by-step answer:
We have two circles whose equations are \[{{x}^{2}}+{{y}^{2}}-2x+8y-\lambda =0\] and \[{{x}^{2}}+{{y}^{2}}+4x+22y+\lambda =0\].
The circumference of one circle is bisected by another circle. We have to find the value of parameter \[\lambda \].
We will begin by finding the centre of the two circles.
We know that the circle of the form \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] has centre at point \[\left( -g,-f \right)\].
Thus, the centre of the circle \[{{x}^{2}}+{{y}^{2}}-2x+8y-\lambda =0\] is \[\left( 1,-4 \right)\].
Similarly, the centre of the circle \[{{x}^{2}}+{{y}^{2}}+4x+22y+\lambda =0\] is \[\left( -2,-11 \right)\].
Now, we will solve the two equations of circles to find the common chord of the given circles.
Subtracting the two equations of circles, we get \[\left( 2x-8y+\lambda \right)+4x+22y+\lambda =0\] as the equation of common chord.
Thus, the equation of the common chord of the two circles is \[6x+14y+2\lambda =0\].
Now, since the circle \[{{x}^{2}}+{{y}^{2}}-2x+8y-\lambda =0\] is bisected by another circle, the centre of first circle must lie on the common chord of the two circles.
Hence, the point \[\left( 1,-4 \right)\] passes through the line \[6x+14y+2\lambda =0\].
Substituting the value of point in the equation of line, we get \[6\left( 1 \right)+14\left( -4 \right)+2\lambda =0\].
On further solving, we get \[6-56+2\lambda =0\].
Thus, we have \[2\lambda =50\].
Hence, we get \[\lambda =25\].
\[\lambda =25\] is the required answer.

Note: It’s necessary to know the fact that if the circumference of one circle is bisected by another circle, then the centre of the first circle lies on the common chord of the two circles.Students should know fact that to find the equation of the common chord of two intersecting circles we will subtract the equation (ii) from the equation (i).