The two circles x²+y²-4 x-2 y=0 and x²+y²-6 y+4=0

Correct Answer is : intersect each other

Correct Answer is : intersect each other

Test Index

UPSC NDA Math Model Paper 5

Question : 119 of 120

Marks: +1, -0

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

x^{2}+y^{2}-6 y+4=0

Solution:

Given circles are

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

and

x^{2}+y^{2}-6 y+4=0

Centre of

1^{\text{st}}

circle

=C_{1}=(2,1)

Radius of

1^{\text{st}}

circle

=r_{1}=\sqrt{(2^{2}+1^{2})}=\sqrt{5}

Centre of

2^{\text{nd}}

circle

=C_{2}=(0,3)

Radius of

1^{\text{st}}

circle

=r_{2}=\sqrt{(0^{2}+3^{2}-4)}=\sqrt{5}

Now,

C_{1}\ C_{2}=\sqrt{((0-2)^{2}+(3-1)^{2})}

=\sqrt{(2^{2}+2^{2})}=\sqrt{8}=2\sqrt{2}

And

r_{1}+r_{2}=\sqrt{5}+\sqrt{5}=2\sqrt{5}

Here

r_{1}+r_{2}>C_{1}C_{2}

Thus, two circles intersect each other.

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