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Circle

Section: Mathematics

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Question : 87 of 89

Marks: +1, -0

If

(a, b)

(c, d)

are the internal and external centres of similitudes of the circles

x^2+y^2+4 x-5=0

x^2+y^2-6 y+8=0

respectively, then

(a+d)(b+c)=

[AP EAPCET 18 May 2024 Shift 1]

4
9
13
22

Solution:

We have, equation of two circlet

x^2+y^2+4 x-5=0

and

x^2+y^2-6 y+8=0

Centre and radius of both circles are

\;c_1= (-2,0, r_1=\sqrt{4+5}=3.

\;c_2=(0,3), r_2=\sqrt{9-8}=1

Let

P

and

Q

be the internal and external centre of similitude, it divides

C_1 C_2

is ine ratio

3: 1

.

P = \frac{x_2+C_1}{3+1} = \left( \frac{-2}{4}, \frac{9}{4} \right) = \left( \frac{-1}{2}; \frac{9}{4} \right)

Here,

a = \frac{-1}{2}

and

b = \frac{9}{4}

Q = \frac{3 C_2-C_1}{C_1-C_2} = \left( \frac{2}{2}, \frac{9}{2} \right) = \left( 1, \frac{9}{2} \right)

Here,

C=1

and

d = \frac{9}{2}

\text{ Hence, } (a+\Delta)(b+c) = \left( \frac{-1}{2} + \frac{9}{2} \right) \left( \frac{9}{4} + 1 \right) \\ = \frac{4 \times 13}{4} = 13

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