JIPMAT 2022 (QA) - The equation of the circle which passes through the point (4,5) and its centre at (2,2) is | PYQs + Solutions

JIPMAT 2022

Geometry

Circles

Medium

The equation of the circle which passes through the point $(4,5)$ and its centre at $(2,2)$ is

(x-2)+(y-2)=13

(x-2)^{2}+(y-2)^{2}=13

x^{2}+y^{2}=13

(x-4)^{2}+(y-5)^{2}=13

Correct Option: 2

The equation of a circle:

(x-a)^{2}+(y-b)^{2}=r^{2}

With centre of

(a, b)

and a radius of

r

Only option (2) matches this, hence you can mark the answer right here.

However, here's the rest of the solution:

\newline

\newline

According to the question,

x=4,y=5

Go by the options, look at Option 2 and:

(4-2)^2+(5-2)^2=2^2+3^2=4+9=13 =RHS

Hence it's the answer.

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