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Question

A circle whose centre is in 2nd quadrant touches the line y=x at a point P (in 3rd quadrant) such that OP=42 unit, where O is the origin. The length of its chord on the line x+y=0 is 62 units. If end points of chord is A,B and centre is C, then sum of ordinates of P,A,B,C is

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Solution

Given OP=42 unit
So possible coordinates of P is (4,4)
and centre lies in 2nd quadrant
So poosible diagram for the question is
Now OP2=OAOB
32=OA(AB+OA)32=OA(62+OA)OA=22 units
OB=82 units
B(8,8),A(2,2)
Now perpendicular bisector of AB is
xy+10=0(1)
and equation of line OP and passing through P is
x+y+8=0(2)
Intersection of (1) and (2) will give the centre of the circle as (9,1)
Hence the required sum =8+2+14=7

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