The point of contact of the circles x²+y²+2 x+2 y+1=0 and x²+y²-2 x+2 y+1=0 is

Correct Answer is : (0,−1)(0,-1)(0,−1)

Correct Answer is : (0,−1)(0,-1)(0,−1)

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AP EAMCET Engineering 2011 Solved Paper

Section: Mathematics

Question : 151 of 160

Marks: +1, -0

x^{2}+y^{2}+2 x+2 y+1=0

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

Solution:

Given equation of circles are

S_{1} \equiv x^{2}+y^{2}+2 x+2 y+1=0

and

S_{2} \equiv x^{2}+y^{2}-2 x+2 y+1=0

In circle

S_{1}

, centre

c_{1}=(-1,-1)

, radius

r_{1}=\sqrt{1+1-1}=1

In circle

S_{2}

, centre

C_{2}=(1,-1)

, radius

r_{2}=\sqrt{1+1-1}=1

Now, distance between two centre

C_{1} C_{2} =\sqrt{(-1-1)^{2}+0}

=2

Here, we see that

C_{1} C_{2}=r_{1}+r_{2}

i.e., both circle touch externally.

So, point of contact of circle

S_{1}

and

S_{2}

=

Mid-point of

C_{1}

and

C_{2}

=\left(\frac{-1+1}{2}, \frac{-1-1}{2}\right)=(0,-1)

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