In the given figure, the equation of the larger circle is x²+y²+4y-5=0 and the distance between centres is 4. Then the equation of smaller circle is

Correct Answer is : (x−7)2+(y−1)2=1(x-{7})²+(y-1)²=1(x−7)2+(y−1)2=1

Correct Answer is : (x−7)2+(y−1)2=1(x-{7})²+(y-1)²=1(x−7)2+(y−1)2=1

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BITSAT 2020 Paper

Section: Mathematics

Question : 108 of 150

Marks: +1, -0

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

4.

Solution:

We have

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

Its centre is

C_{1}(0,-2)

r_{1}=\sqrt{4+5}=3.

Let

C_{2}(h, k)

be the centre of the smaller circle and its radius

r_{2}

Then

C_{1}C_{2}=4

\Rightarrow\sqrt{h^{2}+(k+2)^{2}}=3+r_{2}=4

...(1)

\Rightarrow r_{2}=1

But

k=r_{2}=1

[it touches

x

-axis

\therefore

From

eq(1), 4=\sqrt{h^{2}+(1+2)^{2}}

\Rightarrow 16=h^{2}+9 \Rightarrow h^{2}=7

\Rightarrow h=\pm\sqrt{7}

Since

h>0

\therefore h=\sqrt{7}

Hence, required circle is

(x-\sqrt{7})^{2}+(y-1)^{2}=1

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