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Circle

Section: Mathematics

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Question : 22 of 89

Marks: +1, -0

If

P(\alpha, \beta)

is the radical centre of the circles

S \equiv x^2 + y^2 + 4x + 7 = 0

S' \equiv 2x^2 + 2y^2 + 3x + 5y + 9 = 0

S'' \equiv x^2 + y^2 + y = 0

, then the length of the tangent drawn from P to

S' = 0

[AP EAPCET 23 May 2025 S2]

5
8
4
2

Solution:

The given equations are

s: x^2 + y^2 + 4x + 7 = 0

s': 2x^2 + 2y^2 + 3x + 5y + 9 = 0

\Rightarrow x^2 + y^2 + \frac{3}{2}x + \frac{5}{2}y + \frac{9}{2} = 0

s'': x^2 + y^2 + y = 0

Radical axis of

s

and

s': s - s' = 0

\Rightarrow 4x - \frac{3}{2}x - \frac{5}{2}y + 7 - \frac{9}{2} = 0

\Rightarrow \frac{5}{2}x - \frac{5}{2}y + \frac{5}{2} = 0

\Rightarrow x - y + 1 = 0 \cdots (i)

Radical axis of

s

and

s'': s - s'' = 0

\Rightarrow 4x - y + 7 = 0 \cdots (ii)

From Eqs. (i) and (ii), we have

x = -2, y = -1

∴ Radical centre is

P(-2, -1)

Length of tangent from

P

to

s'

= \sqrt{(-2)^2 + (-1)^2 + \frac{3}{2}(-2) + \frac{5}{2}(-1) + \frac{9}{2}}

= \sqrt{4 + 1 - 3 + 2} = \sqrt{4} = 2

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