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Circle

Section: Mathematics

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Question : 25 of 89

Marks: +1, -0

Let

\theta

be the angle between the circles

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

S^1 \equiv x^2+y^2-6x-8y+9=0

c

is an integer and

\cos \theta = \frac{5}{16}

then the radius of the circle

S=0

[AP EAPCET 22 May 2025 S1]

2
4
3
1

Solution:

Given equation of circles,

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

S^{'} = x^2+y^2-6x-8y+9=0

(g_1, f_1) \equiv -1, 1, c_1=c

(g_2, f_2) \equiv (3,4), c_2=9

Angle between circle

S

and

S^{'}

be

\theta

\cos \theta = \frac{2(g_1 g_2 + f_1 f_2) - (c_1 + c_2)}{2 \sqrt{g_1^2 + f_1^2 - c_1} \sqrt{g_2^2 + f_2^2 - c_2}}

\Rightarrow \cos \theta = \frac{2(-3+4)-c-9}{2\sqrt{2-c}\sqrt{25-9}}

\Rightarrow \frac{5}{16} = \frac{-c-7}{2\sqrt{2-c}(4)}

\Rightarrow 5 \times \sqrt{2-c} = -2(c+7)

\Rightarrow 25(2-c) = 4(c^2+14c+49)

\Rightarrow 4c^2+81c+146=0

\Rightarrow c=-2 \; \frac{-146}{8}

\therefore c=-2

\text{Radius} = \sqrt{1+1+2} = 2

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