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Q. A circle passes through the point `(a,b)` and cuts the circle `x^2 + y^2 = 4`...

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user8235304 | Student, Grade 11 | Valedictorian

Posted July 14, 2013 at 5:54 PM via web

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Q. A circle passes through the point `(a,b)` and cuts the circle `x^2 + y^2 = 4` orthogonally, then the locus of its centre is:

A) `(3x - 1)^2 + (3y)^2 = a^2 + b^2`

B) `(3x + 1)^2 + (3y)^2 = a^2 + b^2`

C) `(3x + 1)^2 + (3y)^2 = a^2 - b^2`

D) `a^2 + b^2 - 2ax - 2by + 4=0`

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aruv | High School Teacher | Valedictorian

Posted July 14, 2013 at 6:50 PM (Answer #1)

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Correct answer is D.

becuase

Let `(x_1,y_1) `  be the centre of the circle.

Thus radii of circles are

`r_1=sqrt((x_1-a)^2+(y_1-b)^2)`

`r=sqrt(4)=2`

distance between centres  is

`d=sqrt(x^2_1+y^2_1)`

Since circles intersect orthogonally therefore

`r^2_1+r^2=d^2`

`(x_1-a)^2+(y_1-b)^2+4=x^2_1+y^2_1`

`a^2+b^2-2ax_1-2by_1+4=0`

Thus locus of centre is

`a^2+b^2-2ax-2by+4=0`

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