Question

Find the equation of the circle which passes through the points (3, 4) and (- 1, 2) and
whose centre lies on the line x - y = 4
.

Answer 1

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

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

So, equation of circle is given by

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

Now, Equation (1), passes through ( 3, 4 ).

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

Now, Equation (1), passes through ( - 1, 2 ), So,

$\rm :\longmapsto\: {( - 1 - h)}^{2} + {(2 - k)}^{2} = {r}^{2} - - - (3)$

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

$\rm :\longmapsto\: {( - 1 - h)}^{2} + {(2 - k)}^{2} = {(3 - h)}^{2} + {(4 - k)}^{2}$

$\rm :\longmapsto\:1 + {h}^{2} + 2h + 4 + {k}^{2} - 4k = 9 + {h}^{2} - 6h + 16 + {k}^{2} - 8k$

$\rm :\longmapsto\:5 + 2h - 4k = 25 - 6h - 8k$

$\rm :\longmapsto\:2h - 4k + 6h + 8k = 25 - 5$

$\rm :\longmapsto\: 8h + 4k = 20$

On dividing by 4, we get

$\bf :\longmapsto\: 2h + k = 5 - - - - (4)$

Now, given that centre ( h, k ) lies on the line x - y = 4

So,

$\bf :\longmapsto\: h - k = 4 - - - - (5)$

On adding equation (4) and (5), we get

$\rm :\longmapsto\:3h = 9$

$\rm :\implies\:h \: = \: 3 - - - (6)$

On substituting h = 3, in equation (5), we get

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

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

$\rm :\longmapsto\: - k = 1$

$\bf\implies \:k \: = \: - \: 1 - - - - (7)$

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

$\rm :\longmapsto\: {(3 - 3)}^{2} + {(4 + 1)}^{2} = {r}^{2}$

$\rm :\longmapsto\: {0}^{2} + {5}^{2} = {r}^{2}$

$\rm :\longmapsto\: {5}^{2} = {r}^{2}$

$\bf\implies \:r \: = \: 5 - - - - (8)$

So, on substituting the values of h, k and r in equation (1), we get

$\rm :\longmapsto\: {(x - 3)}^{2} + {(y + 1)}^{2} = {5}^{2}$

$\rm :\longmapsto\: {x}^{2} + 9 - 6x + {y}^{2} +1 + 2y = 25$

$\bf :\longmapsto\: {x}^{2} + {y}^{2} - 6x + 2y - 15 = 0$

Answer 2

Answer:

$\huge \fbox \red{given}$

Find the equation of the circle which passes through the points (3, 4) and (- 1, 2) and

whose centre lies on the line x - y = 4.

$\huge \mathfrak \red{solution}$

$\large\underline{\sf{Solution-}} </p><p>$

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

So, equation of circle is given by

\

$\rm :\longmapsto\: {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} - - - (1):⟼(x−h) </p><p>2</p><p> +(y−k) </p><p>2</p><p> =r </p><p>2</p><p> −−−(1)$

Now, Equation (1), passes through ( 3, 4 ).

$</p><p>\rm :\longmapsto\: {(3 - h)}^{2} + {(4 - k)}^{2} = {r}^{2} - - - (2):⟼(3−h) </p><p>2</p><p> +(4−k) </p><p>2</p><p> =r </p><p>2</p><p> −−−(2)$

Now, Equation (1), passes through ( - 1, 2 ), So,

$\rm :\longmapsto\: {( - 1 - h)}^{2} + {(2 - k)}^{2} = {r}^{2} - - - (3):⟼(−1−h) </p><p>2</p><p> +(2−k) </p><p>2</p><p> =r </p><p>2</p><p> −−−(3)$

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

$</p><p>\rm :\longmapsto\: {( - 1 - h)}^{2} + {(2 - k)}^{2} = {(3 - h)}^{2} + {(4 - k)}^{2}:⟼(−1−h) </p><p>2</p><p> +(2−k) </p><p>2</p><p> =(3−h) </p><p>2</p><p> +(4−k) </p><p>2$

$\rm :\longmapsto\:1 + {h}^{2} + 2h + 4 + {k}^{2} - 4k = 9 + {h}^{2} - 6h + 16 + {k}^{2} - 8k:⟼1+h </p><p>2</p><p> +2h+4+k </p><p>2</p><p> −4k=9+h </p><p>2</p><p> −6h+16+k </p><p>2</p><p> −8k$

$\rm :\longmapsto\:5 + 2h - 4k = 25 - 6h - 8k:⟼5+2h−4k=25−6h−8k$

$\rm :\longmapsto\:2h - 4k + 6h + 8k = 25 - 5:⟼2h−4k+6h+8k=25−5$

$\rm :\longmapsto\: 8h + 4k = 20:⟼8h+4k=20$

On dividing by 4, we get

$</p><p>\bf :\longmapsto\: 2h + k = 5 - - - - (4):⟼2h+k=5−−−−(4)$

Now, given that centre ( h, k ) lies on the line x - y = 4

So,

$\bf :\longmapsto\: h - k = 4 - - - - (5):⟼h−k=4−−−−(5)$

On adding equation (4) and (5), we get

$\rm :\longmapsto\:3h = 9:⟼3h=9$

\rm :\implies\:h \: = \: 3 - - - (6):⟹h=3−−−(6)

On substituting h = 3, in equation (5), we get

\rm :\longmapsto\:3 - k = 4:⟼3−k=4

\rm :\longmapsto\: - k = 4 - 3:⟼−k=4−3

\rm :\longmapsto\: - k = 1:⟼−k=1

\bf\implies \:k \: = \: - \: 1 - - - - (7)⟹k=−1−−−−(7)

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

\rm :\longmapsto\: {(3 - 3)}^{2} + {(4 + 1)}^{2} = {r}^{2}:⟼(3−3)

2

+(4+1)

2

=r

2

\rm :\longmapsto\: {0}^{2} + {5}^{2} = {r}^{2}:⟼0

2

+5

2

=r

2

\rm :\longmapsto\: {5}^{2} = {r}^{2}:⟼5

2

=r

2

\bf\implies \:r \: = \: 5 - - - - (8)⟹r=5−−−−(8)

So, on substituting the values of h, k and r in equation (1), we get

\rm :\longmapsto\: {(x - 3)}^{2} + {(y + 1)}^{2} = {5}^{2}:⟼(x−3)

2

+(y+1)

2

=5

2

\rm :\longmapsto\: {x}^{2} + 9 - 6x + {y}^{2} +1 + 2y = 25:⟼x

2

+9−6x+y

2

+1+2y=25

\bf :\longmapsto\: {x}^{2} + {y}^{2} - 6x + 2y - 15 = 0:⟼x

2

+y

2

−6x+2y−15=0