Equation of the circle with AB as its diameter is [2010]

Question

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 $B$ . [2010]

Q. Equation of the circle with $A B$ as its diameter is

Smart Achievers · Accepted Answer

(1)

Given a circle

$x^{2} + y^{2} - 8 x = 0 ...(i)$

and a hyperbola $4 x^{2} - 9 y^{2} - 36 = 0 ...(ii)$

To find their point of intersection, substitute the value of $y^{2}$ from equation (i) in equation (ii), we get

$4 x^{2} - 9 (8 x - x^{2}) = 36$ $\Rightarrow 13 x^{2} - 72 x - 36 = 0$

$\Rightarrow x = 6, - \frac{6}{13}$ $\Rightarrow y^{2} = 12, \frac{- 48}{13} - \frac{36}{169} (not possible)$

$∴$ $(6, 2 \sqrt{3}) and (6, - 2 \sqrt{3}) are points of intersection.$

$∴$ $Equation of required circle is$

$(x - 6) (x - 6) + (y - 2 \sqrt{3}) (y + 2 \sqrt{3}) = 0$

$\Rightarrow x^{2} + y^{2} - 12 x + 24 = 0$

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