Q. 104.0( 119 Votes )

# Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.

Answer :

Let the equation of the required circle be (x – h)^{2}+ (y – k)^{2} =r^{2}

Since, the circle passes through points (4,1) and (6,5),

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

(6– h)^{2}+ (5 – k)^{2} =r^{2} ..................(2)

Since, the centre (h,k) of the circle lies on line 4x+y = 16,

4h + k =16..................... (3)

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

(4 – h)^{2}+ (1 – k)^{2} =(6 – h)^{2} + (5 – k)^{2}

16 – 8h + h^{2} +1 -2k +k^{2} = 36 -12h +h^{2}+15 – 10k + k^{2}

16 – 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k =11................ (4)

On solving equations (3) and (4), we obtain h=3 and k= 4.

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

(4 – 3)^{2}+ (1 – 4)^{2} =r^{2}

(1)^{2} + (-3)^{2} = r^{2}

1+9 = r^{2}

r =

Thus, the equation of the requires circle is

(x – 3)^{2}+ (y – 4)^{2} =

⇒ x^{2} -6x + 9 + y^{2} – 8y + 16 =10

**⇒ x ^{2} + y^{2} -6x – 8y = 15 =0**

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