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BITSAT 2011 Paper

Section: Mathematics

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Question : 98 of 150

Marks: +1, -0

For what value of k the circles

x^2 + y^2

+ 5x + 3y + 7 = 0 and

x^2 + y^2

– 8x + 6y + k = 0 cuts orthogonally

16
–18
– 13
– 10

Solution:

Let the two circles be

x^2 + y^2 + 2g_1 x + 2f_1y + c_1

= 0 and

x^2 + y^2 + 2g_2 x + 2f_2y + c_2

= 0

where

g_1

= 5/2,

f_1

= 3/2,

c_1

=7,

g_2

=–4,

f_2

=3 and

c_2

= k

If the two circles intersects orthogonally, then

2

(g_1g_2 + f_1f_2)

=

c-1+c_2

⇒ 2

\left(-10+\frac{9}{2}\right)

= 7 + k

⇒ 11 = 7 + k ⇒ k = 18

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