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The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question:
Equation of the circle with $A B$ as its diameter is
Options:
The circle $x^2+y^2-8 x=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$.Question:
Equation of the circle with $A B$ as its diameter is
Solution:
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Verified Answer
The correct answer is:
$x^2+y^2-12 x+24=0$
$x^2+y^2-12 x+24=0$
The equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$ and that of circle is
$$
x^2+y^2-8 x=0
$$
For their points of intersection
$$
\frac{x^2}{9}+\frac{x^2-8 x}{4}=1
$$
$$
\begin{array}{ll}
\Rightarrow & 4 x^2+9 x^2-72 x=36 \\
\Rightarrow & 13 x^2-72 x-36=0
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow \quad 13 x^2-78 x+6 x-36=0 \\
& \Rightarrow \quad 13 x(x-6)+6(x-6)=0 \\
& \Rightarrow \quad x=6, x=-\frac{13}{6} \\
& x=-\frac{13}{6} \text { not acceptable } \\
&
\end{aligned}
$$
Now, for $x=6, y=\pm 2 \sqrt{3}$
Required equation is
$$
\begin{aligned}
& (x-6)^2+(y+2 \sqrt{3})(y-2 \sqrt{3})=0 \\
& \Rightarrow \quad x^2-12 x+y^2+24=0 \\
& \Rightarrow \quad x^2+y^2-12 x+24=0
\end{aligned}
$$
$$
x^2+y^2-8 x=0
$$
For their points of intersection
$$
\frac{x^2}{9}+\frac{x^2-8 x}{4}=1
$$
$$
\begin{array}{ll}
\Rightarrow & 4 x^2+9 x^2-72 x=36 \\
\Rightarrow & 13 x^2-72 x-36=0
\end{array}
$$
$$
\begin{aligned}
& \Rightarrow \quad 13 x^2-78 x+6 x-36=0 \\
& \Rightarrow \quad 13 x(x-6)+6(x-6)=0 \\
& \Rightarrow \quad x=6, x=-\frac{13}{6} \\
& x=-\frac{13}{6} \text { not acceptable } \\
&
\end{aligned}
$$
Now, for $x=6, y=\pm 2 \sqrt{3}$
Required equation is
$$
\begin{aligned}
& (x-6)^2+(y+2 \sqrt{3})(y-2 \sqrt{3})=0 \\
& \Rightarrow \quad x^2-12 x+y^2+24=0 \\
& \Rightarrow \quad x^2+y^2-12 x+24=0
\end{aligned}
$$
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