QUESTION

A circle passes through the point (3, 4) and cuts the circle ${x^2} + {y^2} = {a^2}$ orthogonally, the locus of its centre is a straight line. If the distance of this straight line from the origin is 25, then ${a^2}$ =A.250B.225C.100D.25

Hint – Limiting points of the system of co- axial circles are the centres of the point circles belonging to the family. And as we know, every circle passing through limiting points of a coaxial system is orthogonal to all the circles of the system.

So, let there be a circle with radius 0 and passes through (3, 4). Equation of this circle is ${S_1}:{(x - 3)^2} + {(y - 4)^2} = 0$.
We have been given in the question the equation of the circle ${x^2} + {y^2} = {a^2}$.
Let the given circle be ${S_2}:{x^2} + {y^2} - {a^2} = 0$.
Therefore, equation of the radical axis: ${S_1} - {S_2} = 0$
$\Rightarrow - 6x - 8y + 25 + {a^2} = 0$
$\dfrac{{|25 + {a^2}|}}{{\sqrt {36 + 64} }} = 25 \\ \Rightarrow 25 + {a^2} = 250 \\ \therefore {a^2} = 225 \\$
Note – Whenever such types of questions appear then write the equation of given circle and the equation of the circle that cuts the given circle orthogonally is ${S_1}:{(x - 3)^2} + {(y - 4)^2} = 0$, as this circle has radius 0 and passes through (3, 4), as it is a point circle. Using these two equations of circle find the equation of radical axis, which is the required locus and then equate the distance of (0, 0) from the radical axis with 25 to find the value of ${a^2}$.