JEE Advanced · Mathematics · 12. Circle
Paragraph:
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
- A
\(x^2+y^2-12 x+24=0\)
- B
\(x^2+y^2+12 x+24=0\)
- C
\(x^2+y^2+24 x-12=0\)
- D
\(x^2+y^2-24 x-12=0\)
Answer & Solution
Correct Answer
(A)
\(x^2+y^2-12 x+24=0\)
Step-by-step Solution
Detailed explanation
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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