Test Index

KEAM 2011 Math Paper

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Question : 11 of 120

Marks: +1, -0

The sum of the minimum distance and the maximum distance from the point

(4 - 3)

x^2+y^2+4x-10y-7=0

is

20
12
10
16
22

Solution:

Given equation of circle

x^{2}+y^{2}+4 x-10 y-7=0

with centre

=(-2,5)=C

and

radius

=\sqrt{4+25+7}=\sqrt{36}=6

Now,

P C=\sqrt{(4+2)^{2}+(-3-5)^{2}}

=\sqrt{36+64}=\sqrt{100}=10

So, the point lies outside the circle.

Now, maximum distance

=P Q=P C+Q C=10+6=16

and minimum distance

=P S=P C-S C=10-6=4

∴ Sum of maximum and minimum distance

=16+4=20

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