Question

10. Find the equation of the circle passing through the points (4.1) and (6.5) and whose centre is on the line
(2.3) and (-1.1) and whose centre is on the line​

Answer 1

Solution :-

Let assume that centre of circle be (h, k) and radius be 'r'.

So,

Equation of circle is

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

Now, equation (1) passes through (4, 1)

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

Also, equation (1) passes through (6, 5),

$\rm :\longmapsto\: {(6 - h)}^{2} + {(5 - k )}^{2} = {r}^{2} - - - (3)$

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

$\rm :\longmapsto\: {(6 - h)}^{2} + {(5 - k)}^{2} = {(4 - h)}^{2} + {(1 - k)}^{2}$

$\rm \to \: 36+{h}^{2}-12h+25+{k}^{2} - 10k= 16+{h}^{2}-8h+1+{k}^{2}-2k$

$\rm :\longmapsto\:61 - 12h - 10k = 17 - 8h - 2k$

$\rm :\longmapsto\: - 12h - 10k + 8h + 2k = 17 - 61$

$\rm :\longmapsto\: - 4h - 8k = - 44$

$\bf\implies \:h + 2k = 11 - - - (4)$

Now, equation of line passes through two points (2, 3) and (- 1, 1) is

$\rm :\longmapsto\:y - 3 = \dfrac{1 - 3}{ - 1 - 2}(x - 2)$

$\blue{ \quad\boxed{ \quad \because{ \: \bf \:y - y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1) \quad}}}$

$\rm :\longmapsto\:y - 3 = \dfrac{ - 2}{ -3}(x - 2)$

$\rm :\longmapsto\:3y - 9 = 2x - 4$

$\rm :\longmapsto\:2x - 3y = - 5$

As centre (h, k) lies on the line 2x - 3y = - 5

$\rm :\implies\:2h - 3k = - 5$

$\rm :\longmapsto\:2(11 - 2k) - 3k = - 5 \: \: \: \: \: \: \{ \: using \: (4) \}$

$\rm :\longmapsto\:22 - 4k - 3k = - 5$

$\rm :\longmapsto\:22 - 7k \: = - 5$

$\bf\implies \:k = \dfrac{27}{7}$

Put value of k, in equation (4), we get

$\rm :\longmapsto\:h + 2(\dfrac{27}{7} ) = 11$

$\rm :\longmapsto\:h + \dfrac{54}{7} = 11$

$\rm :\longmapsto\:h = 11 - \dfrac{54}{7}$

$\rm :\longmapsto\:h = \dfrac{77 - 54}{7}$

$\bf\implies \:h = \dfrac{23}{7}$

Now, on substituting h and k in equation (2), we get

$\rm :\longmapsto\: {(4 - \dfrac{23}{7})}^{2} + {(1 - \dfrac{27}{7} )}^{2} = {r}^{2}$

$\rm :\longmapsto\: {(\dfrac{28 - 23}{7})}^{2} + {(\dfrac{7 - 27}{7} )}^{2} = {r}^{2}$

$\rm :\longmapsto\: {(\dfrac{5}{7})}^{2} + {(\dfrac{ - 20}{7} )}^{2} = {r}^{2}$

$\rm :\longmapsto\: {r}^{2} = \dfrac{25}{49} + \dfrac{400}{49}$

$\rm :\longmapsto\: {r}^{2} = \dfrac{400 + 25}{49}$

$\rm :\longmapsto\: {r}^{2} = \dfrac{425}{49}$

So,

required equation of circle is obtained by substituting the values of h, k and r in equation (1), we gat

$\rm :\longmapsto\: {(x - \dfrac{23}{7} )}^{2} + {(y - \dfrac{27}{7} )}^{2} = \dfrac{425}{49}$

$\rm :\longmapsto\: {x}^{2} + \dfrac{529}{49} - \dfrac{46}{7}x + {y}^{2} + \dfrac{729}{49} - \dfrac{54}{7}y = \dfrac{425}{49}$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - \dfrac{46}{7}x - \dfrac{54}{7}y + \dfrac{833}{49} = 0$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} - \dfrac{46}{7}x - \dfrac{54}{7}y + 17 = 0$

$\rm :\longmapsto\: {7x}^{2} + {7y}^{2} - 46x - 54y + 119 = 0$