Test Index

TS EAMCET 4 May 2018 Shift 2 Solved Paper

Section: Mathematics

© examsnet.com

Question : 50 of 160

Marks: +1, -0

If

\quad

x^{2}+y^{2}+2kx-4y+1=0

x^{2}+y^{2}-8x-12y+43=0

touch each other then

k=

2
1
−1
−2

Solution:

It is given that,

r_{1}=\sqrt{k^{2}+4-1}

=\sqrt{k^{2}+3}

And,

r_{2}=\sqrt{16+36-43}

=\sqrt{9}

=3

The common tangent of above two is given by,

S_{1}-S_{2}=0

So,

x^{2}+y^{2}+2kx-4y+1-x^{2}

-y^{2}+8x+12y-43=0

(2k+8)x+8y-42=0

The perpendicular drawn on the circle is given by,

\frac{(2k+8)4+48-42}{\sqrt{(2k+8)^{2}+64}}=3

8k+38=3\sqrt{4k^{2}+32k+128}

[8k+38]^{2}=9(4k^{2}+32k+128)

k=-1

© examsnet.com

Go to Question:

Prev Question

Next Question