If y=2x is a chord of the circle x^2+y^2-10x=0, then the equation of the circle of which this chord is a diameter, is x^2+y^2- Ax- By=0. Find A+B.

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Correct answer: 6. 6JEE Main 2020 Family of circles passing from points of intersection of circle and chord are x^2+y^2-10x+ (y-2x )=0 Its center is ( (5+ ) ,- 2) It lies on y-2x=0. i.e. - 2-10-2 =0 ⇒-5 =20 ⇒ =-4 Hence equation of circle is x^2+y^2-2x-4y=0. A=2 ,B=4 A+B=6 Trick: Since y=2x is the line passing through origin and through I and III quadrant. Also centre of the given circle lies on positive x -axis, therefore, the centre of required circle will be in I quadrant and must satisfy the given line y=2x and its equation is given by x^2+y^2-2x-4y=0

$If y = 2 x is a chord of the circle x^{2} + y^{2} - 10 x = 0, then the equation of the circle of which this chord is a diameter, is x^{2} + y^{2} - Ax - By = 0 . Find A + B .$

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