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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 $a^2$ is equal to
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The correct answer is:
$225$
Since, the circle passes through $(3,4)$ and cuts the circle $x^2+y^2=a^2$ orthogonally.
$\therefore \quad(x-3)^2+(y-4)^2=0$
Also, $\quad x^2+y^2-a^2=0$
$\therefore$ Fquation of radical axis,
$\begin{aligned} & (x-3)^2+(y-4)^2-\left(x^2+y^2-a^2\right)=0 \\ \Rightarrow & x^2-6 x+9+y^2+16-8 y-x^2 \\ & -y^2+a^2=0 \\ \Rightarrow & -6 x-8 y+25+a^2=0 \\ \Rightarrow & 6 x-8 y-25-a^2=0\end{aligned}$
Now, the distance from $(0,0)$ to Eq. (iii), we get
$\begin{aligned} & \frac{\left|6(0)-8(0)-25-a^2\right|}{\sqrt{36+64}}=25 \\ & \Rightarrow \quad \frac{25+a^2}{10}=25 \\ & \Rightarrow \quad 25+a^2=250 \\ & \Rightarrow \quad a^2=225 \\ & \end{aligned}$
$\therefore \quad(x-3)^2+(y-4)^2=0$
Also, $\quad x^2+y^2-a^2=0$
$\therefore$ Fquation of radical axis,
$\begin{aligned} & (x-3)^2+(y-4)^2-\left(x^2+y^2-a^2\right)=0 \\ \Rightarrow & x^2-6 x+9+y^2+16-8 y-x^2 \\ & -y^2+a^2=0 \\ \Rightarrow & -6 x-8 y+25+a^2=0 \\ \Rightarrow & 6 x-8 y-25-a^2=0\end{aligned}$
Now, the distance from $(0,0)$ to Eq. (iii), we get
$\begin{aligned} & \frac{\left|6(0)-8(0)-25-a^2\right|}{\sqrt{36+64}}=25 \\ & \Rightarrow \quad \frac{25+a^2}{10}=25 \\ & \Rightarrow \quad 25+a^2=250 \\ & \Rightarrow \quad a^2=225 \\ & \end{aligned}$
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