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Question

A circle touches the line y=x at a point P such that OP=42, where O is the origin. The circle makes an intercept of 62 units on line x+y=0. Then the equation of the circle(s) is/are

A
(x9)2+(y+1)250=0
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B
(x1)2+(y+9)250=0
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C
(x+1)2+(y9)250=0
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D
(x+9)2+(y1)250=0
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Solution

The correct option is D (x+9)2+(y1)250=0
Let AB be the length of chord intercepted by circle on y+x=0
Let CM be perpendicular to AB from centre C(h,k).

Also, yx=0 and y+x=0 are perpendicular to each other.
OPCM is rectangle.
CM=OP=42.
and AB=62
Now AB=2AM=2r2(CM)2
r=52

Again y=x touches the circle.
CP=r
hk2=52hk=±10 (1)
Also, CM=42
h+k2=42h+k=±8 (2)
Solving equations (1) and (2), we get the possible centres as (9,1),(1,9),(1,9),(9,1)

Possible circles are
(x9)2+(y+1)250=0
(x1)2+(y+9)250=0
(x+1)2+(y9)250=0
(x+9)2+(y1)250=0

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