Question

Find the equation of the circle passing through the points (0, 2), (3, 0)
and (3, 2).
​

Answer 1

Three points on the circle are:−A≡(0,2),B≡(3,0),C≡(3,2)

Let P≡(x,y) be its centre, then ∣PA∣=∣PB∣=∣PC∣

⇒

(x−0)

2

+(y−2)

2

=

(x−3)

2

+(y−0)

2

=

(x−3)

2

+(y−2)

2

⇒

x

2

+y

2

+4−4y

=

x

2

+9−6x+y

2

=

x

2

+9−6x+y

2

+4−4y

⇒4−4y=9−6x=9−6x+4−4y

Taking first and third parts of the above equation,

4−4y=9+4−6x−4y⇒6x=9⇒x=

6

9

=

2

3

Taking second and third parts of the above equation,

9−6x=9+4−6x−4y⇒4y=4⇒y=1∴P≡(

2

3

,1)

So, radius r=PA=

(

2

3

)

2

+(1−2)

2

=

4

9

+1

=

2

5

Thus, equation of the circle ;_(x−

2

3

)

2

+(y−1)

2

=(

2

5

)

2

⇒x

2

+y

2

−3x−2y+1−

4

5

+

4

9

=0⇒x

2

+y

2

−3x−2y=0