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KEAM 2014 Math Paper

Question : 113 of 120

Marks: +1, -0

The centre of the circle whose radius is 5 and which touches the circle

x^2+y^2-2x-4y-20=0

(5,5)

(10 ,5)
(5 , 8 )
(5 , 10)
(8 , 9 )
(9 , 8)

Validate

Solution:

The centre and radius of given circle

x^{2}+y^{2}-2 x-4 y-20=0

C_{1}(1,2)

and

r=\sqrt{1^{2}+2^{2}+20}=5

since, the radii of the two circles are equal. therefore the two circles will touch externally and the point of contact will be the mid-point between two centres.

Let coordinate of the centre of required circle

\text{be}(h, k)

\therefore \;\; \;\frac{h+1}{2}=5

and

\;\frac{k+2}{2}=5

\Rightarrow \;\; h=9

and

k=8

Hence, the centre is (9,8)

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