IIT-JEE 2010 · Paper 1 · Mathematics Q46

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MathematicsSingle correct+3 / -1Question 46 of 84

Passage

Paragraph for Questions 45 to 46

The circle $x^2+y^2-8x=0$ and hyperbola $\dfrac{x^2}9-\dfrac{y^2}4=1$ intersect at the points A and B.

Equation of the circle with AB as its diameter is

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

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

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

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

Answer and worked solution

Official answer: A

Intersections. From the hyperbola $y^2=\frac49x^2-4$ ; substitute into $x^2+y^2=8x$ :
$\frac{13}9x^2-8x-4=0\implies13x^2-72x-36=0\implies x=6\text{ or }-\frac6{13}.$
On the hyperbola $|x|\ge3$ , so $x=6$ , $y^2=12$ : $A(6,2\sqrt3)$ , $B(6,-2\sqrt3)$ .

Circle on AB. Centre $(6,0)$ , radius $2\sqrt3$ :
$(x-6)^2+y^2=12\iff x^2+y^2-12x+24=0.$

(A)

Topics:Equation of a circle Hyperbola

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IIT-JEE 2010 · Paper 1, Mathematics question 46