Test Index

BITSAT 2021 Paper

Section: Mathematics

Question : 131 of 150

Marks: +1, -0

y=5

(-1,2)

(1,2)

Solution:

The centre of the circle is on the perpendicular bisector of the line joining

(-1,2)

and

(1,2)

, which is the y - axis. The ordinate of the centre is given by

(5-y)^{2}=1+(y-2)^{2}

⇒

y=\frac{10}{3}

Hence, eq. of the circle is :

x^{2}+\left(y-\frac{10}{3}\right)^{2}=\left(\frac{5}{3}\right)^{2}

⇒

9 x^{2}+9 y^{2}-60 y+75=0

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