Paragraph: The circle x^2+y^2-8 x=0 and hyperbola (x^2)/(9)-(y^2)/(4)=1 intersect at the points A and B . Question: Equation of the circle with A B as its diameter is

Correct answer: A. x^2+y^2-12 x+24=0 The equation of the hyperbola is (x^2)/(9)-(y^2)/(4)=1 and that of circle is x²+y²-8 x=0 For their points of intersection (x^2)/(9)+(x^2-8 x)/(4)= 1 array ll & 4 x²+9 x²-72 x=36 & 13 x²-72 x-36=0 array aligned & 13 x²-78 x+6 x-36=0 & 13 x(x-6)+6(x-6)=0 & x=6, x=-(13)/(6) & x=-(13)/(6) not acceptable & aligned Now, for x=6, y=± 2 3 Required equation is aligned & (x-6)^2+(y+2 3 )(y-2 3 )=0 & x²-12 x+y²+24=0 & x²+y²-12 x+24=0 aligned

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Circle

JEE Advanced 2010 · Paper 1 · Mathematics · Circle

Paragraph: The circle x^2+y^2-8 x=0 and hyperbola...

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

Options

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 $x^2+y^2-24 x-12=0$

Correct answer

A. $x^2+y^2-12 x+24=0$

Step-by-step solution

The equation of the hyperbola is

\frac{x^2}{9}-\frac{y^2}{4}=1

and that of circle is x²+y²-8 x=0 For their points of intersection

\frac{x^2}{9}+\frac{x^2-8 x}{4}=

\begin{array}{ll} \Rightarrow& 4 x²+9 x²-72 x=36 \\ \Rightarrow & 13 x²-72 x-36=0 \end{array}

\begin{aligned}& \Rightarrow \quad 13 x²-78 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 x²-12 x+y²+24=0 \\ & \Rightarrow \quad x²+y²-12 x+24=0 \end{aligned}

Practice Circle on Quantrex Academy →

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