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} =r^{2}

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

(2 – h)^{2}+ (3 – k)^{2} =r^{2} .................(1)

(-1 – h)^{2}+ (1– k)^{2} =r^{2} ..................(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 + h^{2} +9 -6k +k^{2} = 1 + 2h +h^{2}+1 – 2k + k^{2}

⇒ 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

+= r^{2}

+= r^{2}

+ = r^{2}

+ = r^{2}

Thus, the equation of the required circle is

+=

+=

⇒ 4x^{2} -28x + 49 +4y^{2} + 20y + 25 =130

⇒ 4x^{2} +4y^{2} -28x + 20y - 56 =0

⇒ 4(x^{2} +y^{2} -7x + 5y – 14) = 0

⇒ x^{2}+y^{2}-7x + 5y– 14 =0

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