Question:
Two circles $ x^2 + y^2 = 6$ and $ x^2 + y^2 - 6x + 8 = 0$ are given. Then the equation of the circle passing through their points of intersection and the point (1, 1) is
Two circles $ x^2 + y^2 = 6$ and $ x^2 + y^2 - 6x + 8 = 0$ are given. Then the equation of the circle passing through their points of intersection and the point (1, 1) is
Updated On: Jun 14, 2022
- $ x^2 + y^2 - 6x + 4 = 0$
- $ x^2 + y^2 - 3x + 1 =0$
- $ x^2 + y^2 - 4y + 2 = 0 $
- none of these.
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The Correct Option is B
Solution and Explanation
The required equation of circle is, $S_1 + \lambda (S_2 ~ S_1 ) = 0$
$\therefore \, \, \, \, \, \, \, \, \, \, \, ( x^2 + y^2 -6) + \lambda (-6x +14)=0$
Also, passing through (1, 1).
$\Rightarrow \, \, \, \, \, \, \, \, -4 + \lambda (8) = 0 \Rightarrow \lambda = \frac{1}{2}$
$\therefore \, Required equation of circle is$
$ \, \, \, \, \, \, \, \, \, \, \, x^2 + y^2 - 6 - 3x +7 = 0$
or $ \, \, \, \, \, \, \, \, \, \, \, x^2 + y^2 - 3x + 1 = 0$
$\therefore \, \, \, \, \, \, \, \, \, \, \, ( x^2 + y^2 -6) + \lambda (-6x +14)=0$
Also, passing through (1, 1).
$\Rightarrow \, \, \, \, \, \, \, \, -4 + \lambda (8) = 0 \Rightarrow \lambda = \frac{1}{2}$
$\therefore \, Required equation of circle is$
$ \, \, \, \, \, \, \, \, \, \, \, x^2 + y^2 - 6 - 3x +7 = 0$
or $ \, \, \, \, \, \, \, \, \, \, \, x^2 + y^2 - 3x + 1 = 0$
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