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BITSAT 2024Paper 2

Exam: BITSAT
Year: 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Circle

BITSAT 2024 · Paper 2 · Mathematics · Circle

If p and q be the longest and the shortest distance respectively of...

If p and q be the longest and the shortest distance respectively of the point

(-7,2)

from any point

(\alpha, \beta)

on the curve whose equation is

x^2+y^2-10 x-14 y-51=0

, then G.M. of p If p and q be the longest and the shortest distance respectively of the point (-7,2) from any point (α, β) on the curve

If p and q be the longest and the shortest distance respectively of the point (-7,2) from any point (α, β) on the curve

Options

A $2 \sqrt{11}$
B $5 \sqrt{5}$
C $13$
D $11$

Correct answer

A. $2 \sqrt{11}$

Step-by-step solution

The centre

C

of the circle

=(5,7)

and the radius

=\sqrt{5^2+7^2+51}=5 \sqrt{5}\mathrm{PC}=\sqrt{12^2+5^2}=13 \Rightarrow q=P A=13-5 \sqrt{5}

and

p=P B=13+5 \sqrt{3}

If p and q be the longest and the shortest distance respectively of the point (-7,2) from any point (α, β) on the curve

\therefore

G.M. of

p

and

q=\sqrt{p q}=\sqrt{(13-5 \sqrt{5})(13+5 \sqrt{5})}=\sqrt{169-125}=2 \sqrt{11}

.

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