The locus of the midpoints of the chords...

The locus of the midpoints of the chords of the circle `x^2+y^2-a x-b y=0` which subtend a right angle at `(a/2, b/2)` is (a)`a x+b y=0` (b) `a x+b y=a^2+b^2` (c) `x^2+y^2-a x-b y+(a^2+b^2)/8=0` (d) `x^2+y^2-a x-b y-(a^2+b^2)/8=0`

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The locus of the midpoint of a chord of the circle x^2+y^2=4 which subtends a right angle at the origins is (a) x+y=2 (b) x^2+y^2=1 x^2+y^2=2 (d) x+y=1

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The locus of the mid-point of a chord of the circle x^2 + y^2 -2x - 2y - 23=0 , of length 8 units is : (A) x^2 + y^2 - x - y + 1 =0 (B) x^2 + y^2 - 2x - 2y - 7 = 0 (C) x^2 + y^2 - 2x - 2y + 1 = 0 (D) x^2 + y^2 + 2x + 2y + 5 = 0

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Comprehension (Q. 5 & 6) Let A B be any chord of the circle x^2+y^2-2x-6y-6=0 which subtends a right angle at the point (2, 4). Locus of mid point of A B is x^2+y^2-7y-16=0 x^2+y^2-3x-7y+7=0 x^2+y^2+3x+7y-16=0 x^2+y^2+3x+7y-7=0

Solve: x+y=a+b ; a x-b y=a^2-b^2

AB is any chord of the circle x^2+y^2-6x-8y-11=0 which subtends an angle pi/2 at (1,2) . If locus of midpoint of AB is a circle x^2+y^2-2ax-2by-c=0 ; then find the value of (a+b+c) .

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