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Circle

Section: Mathematics

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Question : 16 of 27

Marks: +1, -0

A circle S passes through the point

(0, 1)

and is orthogonal to the circles

(x-1)^2+y^2=16

x^2+y^2=1

[JEE Adv 2014 P1]

radius of S is $8$
radius of S is $7$
center of S is $(-7,1)$
center of S is $(-8,1)$

Solution:

Given Circles

x^2+y^2-2x-15=0

x^2+y^2-1=0

Radical axis

x+7=0

...(1)

Center of circle lies on (1)

Let the center be (-7,k)

Let equation be

x^2+y^2+14x-2ky+c=0

Orthogonality gives

-14=c-15\Rightarrow c=1

....(2)

(0,1)\rightarrow1-2k+1=0\Rightarrow k=1

Hence radius

=\sqrt{7^2+k^2-c}=\sqrt{49+1-1}=7

Alternate solution

Given circle

x^2+y^2-2x-15=0

x^2+y^2-1=0

Let equation of circle

x^2+y^2+2gx+2fy+c=0

Circle passes through

(0,1)

\Rightarrow 1+2f+c=0

Applying condition of orthogonality

-2g=c-15,0=c-1

\Rightarrow c=1,g=7,f=-1

r=\sqrt{49+1-1}=7;

center

(-7,1)

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