Question

Find the equation of the circle passing through the points (5,2)(7,-4) and whose centre lies on y axis

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Centre of circle lies on y - axis.

Let assume that centre of circle be (0, k) and radius of circle be r units.

So, equation of circle is given by

$\rm \: {(x - 0)}^{2} + {(y - k)}^{2} = {r}^{2}$

$\rm \: {x}^{2} + {(y - k)}^{2} = {r}^{2} - - - (1)$

Now, further given that circle passes through the point (5, 2)

So, it means (5, 2) satisfies equation (1).

$\rm \: {5}^{2} + {(2 - k)}^{2} = {r}^{2} - - - (2)$

Also, given that circle passes through the point (7, - 4).

So, it means (7, - 4) satisfies equation (1).

$\rm \: {7}^{2} + {( - 4 - k)}^{2} = {r}^{2} - - - (3)$

On equating equation (2) and (3), we get

$\rm \: {7}^{2} + {( - 4 - k)}^{2} = {5}^{2} + {(2 - k)}^{2}$

$\rm \: 49 + 16 + {k}^{2} + 8k = 25 + {k}^{2} + 4 - 4k$

$\rm \: 65 + 8k = 29 - 4k$

$\rm \: 12k = - 36$

$\rm\implies \:k \: = \: - 3$

On substituting the value of k in equation (2), we get

$\rm \: {5}^{2} + (2 + 3)^{2} = {r}^{2}$

$\rm\implies \: {r}^{2} = 50$

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

$\rm \: {x}^{2} + {(y + 3)}^{2} = 50$

$\rm \: {x}^{2} + {y}^{2} + 9 + 6y = 50$

$\rm \: {x}^{2} + {y}^{2} + 6y - 41 = 0$

So, the required equation of circle passing through the points (5,2)(7,-4) and whose centre lies on y axis is

$\rm\implies \:\boxed{\sf{ \:\rm \: {x}^{2} + {y}^{2} + 6y - 41 = 0 \: \: }}$

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Formulae Used :-

Equation of circle having centre (h, k) and radius r is

$\boxed{\sf{ \: \: {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} \: \: }}$

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Additional Information :-

1. Equation of circle in general form is

$\boxed{\sf{ \: {x}^{2} + {y}^{2} + 2gx + 2fy + c \: = \: 0 \: }}$

having

$\boxed{\sf{ \:Centre \: = \: ( - g, \: - f) \: \: }}$

and

$\boxed{\sf{ \: \: Radius, \: r \: = \sqrt{ {g}^{2} + {f}^{2} - c} \: \: }}$