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. $250$
• B. $225$
• C. $100$
• D. $25$

Answer 1

Answer: (B)

$225$

Note: Limiting Point of system of co-axial circles are the centers of the point circles belonging to the family.
As we know , every circle passing through limiting points of a coaxial system is orthogonal to all circles of the system.
So, any circle passing through a point and orthogonal to a given circle, will have its center on the radical axis of that point circle and that given circle. (or of the coaxial system)
So, let there be a circle with radius $0$ and passes through $(3,4)$. Equation of this circle $S_1:{(x-3)}^2+{(y-4)}^2=0^2$

Given circle $S_2:x^2+y^2-a^2=0$

$\therefore$ Equation of Radical Axis $:S_1-S_2=0$

$-6x-8y+25+a^2=0$
This is the equation of a radical axis of a co-axial system of circles of which $(3,4)$ is a limiting point. Hence, this will be the required locus.

Distance of $(0,0)$ from this Radical Axis is $25$. So,

$\frac{|25+a^2|}{\sqrt{36+64}}=25$

$\Rightarrow 25+a^2=250$

$\therefore a^2=225$