Question

$\huge{ \mathfrak{ \underline{Question}}}$

Find the equation of the circle passing through the point's (4,1) and (6,5) nd whose center is on the line 4x + y = 16​

Answer 1

Step-by-step explanation:

Let the equation of the 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 equations (1) and (2), we obtain

$(4 – h)^2 + (1 – k)^2 = (6 – h)^2 + (5 – k)^2 \\ ⇒ 16 – 8h + h^2 + 1 – 2k + k^2 \\ = 36 – 12h + h^2 + 25 – 10k + k^2$⇒ 16 – 8h + 1 – 2k = 36 – 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

$(4 – 3)^2 + (1 – 4)^2 = r^2 \\ ⇒ (1)^2 + (– 3)^2 = r^2 \\ ⇒ 1 + 9 = r^2 ⇒ r^2 = 10 \\⇒ = \sqrt{10}$

Thus, the equation of the required circle is

$(x – 3)^2 + (y – 4)^2 \\ = (√10)^2 x^2 – 6x + 9 + y^2 – 8y + 16 \\ = 10 x^2 + y^2 – 6x – 8y + 15 = 0$

Answer 2

Step by step explanation:-

Given points,

(4, 1) , (6, 5)

$equation \: of circle {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2}$

$⇒ \: {(4 - h)}^{2} + {(1 - k)}^{2} = {r}^{2} ....(1)$

$⇒ \: {(6 - h) }^{2} + {(5 - k)}^{2} = {r}^{2} ....(2)$

Solving the above 2 equations, we get,

$⇒ h + 2k = 11....(3)$

$⇒ given, 4h + k = 16....(4)$

Solving the above 2 equation, we get,

⇒ h = 3 , k= 4

Substituting the above value in (1) , we get,

$⇒ {(4 - 3)}^{2} + {(1 - 4)}^{2} = {r}^{2}$

$∴r = \sqrt{10}$

Hence the equation is,

$⇒ {(x - 3)}^{2} + {(y - 4)}^{2} = {( \sqrt{10} )}^{2}$

$⇒ {x}^{2} + {y}^{2} - 6x - 8y + 15 = 0$