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Circle

Section: Mathematics

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Question : 30 of 89

Marks: +1, -0

A circle

S \equiv x^2+y^2-16=0

intersects another circle

S'=0

of radius 5 units such that their common chord is of maximum length. If the slope of that chord is

\frac{3}{4}

, then the centre of such a circle

S'=0

[AP EAPCET 22 May 2025 S1]

$\left(\frac{9}{5}, \frac{12}{5}\right)$
$\left(\frac{5}{9}, \frac{-12}{5}\right)$
$\left(\frac{-9}{5}, \frac{12}{5}\right)$
$\left(\frac{3}{5}, \frac{4}{5}\right)$

Solution:

Given that

S \equiv x^2+y^2-16=0

Let equation of

S' \equiv (x-h)^2+(y-k)^2=5^2

Common chord

S_2-S_1=0

2 x h+2 k y+9=k^2+h^2

Given slope

=\frac{3}{4}

\frac{-2h}{2k}=\frac{3}{4} \Rightarrow h=\frac{-3}{4}k

The chord of maximum length is diameter

\therefore (0,0)

lies on common chord

h^2+k^2=9

\frac{9}{16}k^2+k^2=9 \Rightarrow k=\pm \frac{12}{5}

h=\mp \frac{9}{5}

\text{re} \equiv \left(\frac{-9}{5}, \frac{12}{5}\right), \left(\frac{9}{5}, \frac{-12}{5}\right)

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