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.

Asked by Abhisek | 1 year ago |  78

##### Solution :-

Let us consider the equation of the required circle be

(x – h)2+ (y – k)2 = r2

We know that the circle passes through points (4,1) and (6,5)

So,

(4 – h)+ (1 – k)2 = r2 ……………..(1)

(6– h)2+ (5 – k)2 = r2 ………………(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 + h2 +1 -2k +k2 = 36 -12h +h2+15 – 10k + k2

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 = r2

(1)2 + (-3)2 = r2

1+9 = r2

r =$$\sqrt{10}$$

so now, (x – 3)+ (y – 4)2 = $$(\sqrt{10})^2$$

x2 – 6x + 9 + y2 – 8y + 16 =10

x2 + y2 – 6x – 8y + 15 = 0

The equation of the required circle is x2 + y2 – 6x – 8y + 15 = 0

Answered by Pragya Singh | 1 year ago

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