Q. 54.4( 8 Votes )

Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

Answer :

Let the equation of the required circle be (x – h)2+ (y – k)2 =r2


Since, the circle passes through points (2,3) and (-1,1).


(2 – h)2+ (3 – k)2 =r2 .................(1)


(-1 – h)2+ (1– k)2 =r2 ..................(2)


Since, the centre (h,k) of the circle lies on line x - 3y - 11= 0,


h - 3k =11..................... (3)


From the equation (1) and (2), we obtain


(2 – h)2+ (3 – k)2 =(-1 – h)2 + (1 – k)2


⇒ 4 – 4h + h2 +9 -6k +k2 = 1 + 2h +h2+1 – 2k + k2


⇒ 4 – 4h +9 -6k = 1 + 2h + 1 -2k


⇒ 6h + 4k =11................ (4)

Now multiply (3) by 6 and subtract it from (4) to get,

6h+ 4k - 6(h-3k) = 11 – 66

⇒ 6h + 4k – 6h + 18k = 11 – 66

⇒ 22 k = - 55

⇒ K = -5/2

Put this value in (4) to get,

6h + 4(-5/2) = 11

⇒ 6h – 10 = 11

⇒ 6h = 21

⇒ h = 21/6

⇒ h = 7/2


Thus we obtain h=
and k= .


On substituting the values of h and k in equation (1), we get


+= r2


+= r2


+ = r2


+ = r2



Thus, the equation of the required circle is


+


+= 


4x2 -28x + 49 +4y2 + 20y + 25 =130


4x2 +4y2 -28x + 20y - 56 =0


4(x2 +y2 -7x + 5y – 14) = 0


x2+y2-7x + 5y– 14 =0

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