Question

The equation of the circle passing through the points $(2,3)$ and $(-1,1)$ and whose centre is on the line $x-3 y-11=0$, is $x^2+y^2+a x+b y+d=0$, where $a, b$ and $d$ respectively are

• A. $-7,-5,14$
• B. $5,-7,-14$
• C. $-7,5,-14$
• D. $-5,7,14$

Answer 1

Answer: (C)

Let $(h, k)$ be the centre and $r$ be the radius of the circle.
$\therefore $ Equation of circle
$(x-h)^2+(y-k)^2=r^2$ ...(i)
$\because$ Circle (i) passes through points $(2,3)$ and $(-1,1)$.
$\therefore (2-h)^2+(3-k)^2=r^2$....(ii)
$ (-1-h)^2+(1-k)^2=r^2$....(iii)
From Eqs. (ii) and (iii), we have
$ (2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2$
$ \Rightarrow 4+h^2-4 h+9+k^2-6 k=1+h^2+2 h+1+k^2-2 k $
$ \Rightarrow -4 h-2 h-6 k+2 k=2-4-9$
$\Rightarrow -6 h-4 k=-11 $
$ \Rightarrow 6 h+4 k=11$...(iv)
Also, the centre of circle lies on the line $x-3 y=11$
$\therefore h-3 k=11$....(v)
Multiplying Eq. (v) by 6 and subtracting from Eq. (iv), we have
$ (6 h+4 k)-(6 h-18 k)=11-66$
$22 k=-55$
$ k=-\frac{55}{22}$
$ =-\frac{5}{2} $
Substituting the value of $k$ in Eq. (v), we get
$h-3 \times\left(-\frac{5}{2}\right) =11$
$\Rightarrow h+\frac{15}{2} =11 $
$\Rightarrow h =11-\frac{15}{2}=\frac{7}{2}$
Now, substituting the value of $h$ and $k$ in Eq. (ii), we get
$ \left(2-\frac{7}{2}\right)^2+\left(3+\frac{5}{2}\right)^2 =r^2$
$\Rightarrow \left(-\frac{3}{2}\right)^2+\left(\frac{11}{2}\right)^2 =r^2 $
$\Rightarrow \frac{9}{4}+\frac{121}{4} =r^2$
$\frac{130}{4} =r^2$
Required equation of circle
$\left(x-\frac{7}{2}\right)^2+\left(y+\frac{5}{2}\right)^2=\frac{130}{4} $
$\rightarrow x^2+\frac{49}{4}-7 x+y^2+\frac{25}{4}+5 y-\frac{130}{4}$
$\Rightarrow x^2+y^2-7 x+5 y=\frac{130}{4}-\frac{49}{4}-\frac{25}{4} $
$\Rightarrow x^2+y^2-7 x+5 y=\frac{56}{4} $
$ \Rightarrow x^2+y^2-7 x+5 y=14$
or$ x^2+y^2-7 x+5 y-14=0$
Alternate Method
Let equalion or circle is
$x^2+y^2+2 g x+2 f y+c=0$...(i)
Eq. (i) passes through the points $(2,3)$ and $(-1,1)$ i.e., they will satisfy it.
At point $(2,3)$,
$(2)^2+(3)^2+2 g(2)+2 f(3)+c=0 $
$\Rightarrow 4 g+6 f+c+13=0$...(ii)
$\text { and at point }(-1,1),$
$(-1)^2+(1)^2+2 g(-1)+2 f(1)+c=0 $
$\Rightarrow -2 g+2 f+c+2=0$...(iii)
As centre $(-g,-f)$ lies on the line $x-3 y-11=0$
i.e., $(-g)-3(-f)-11=0$
$\Rightarrow -g+3 f-11=0$...(iv)
Now, subtract Eq. (iii) from Eq. (ii), we get
$6 g+4 f+11=0$
From Eq. (iv), $g=3 f-11$ put in Eq. (v), we get
$ 6(3 f-11)+4 f+11=0 $....(v)
$\Rightarrow 18 f-66+4 f+11=0 $
$\Rightarrow 22 f-55=0 $
$\Rightarrow f=\frac{55}{22} \Rightarrow f=\frac{5}{2}$
From Eq. (iv) we get,
$g=3 \times \frac{5}{2}-11=\frac{15-22}{2}=\frac{-7}{2}$
From Eq. (ii) we get,
$4 \times\left(-\frac{c}{2}\right)+6\left(\frac{3}{2}\right)+c+13=0$
$\Rightarrow -14+15+c+13 =0$
$\Rightarrow c+14 =0 $
$\Rightarrow c =-14$
Put the values of $g, f$ and $c$ in Eq. (i) to get the required equation of circle
$ x^2+y^2+2 x\left(-\frac{7}{2}\right)+2 y\left(\frac{5}{2}\right)-14=0 $
$ \Rightarrow x^2+y^2-7 x+5 y-14=0$
Comparing above with $x^2+y^2+a x+b y+d$, we get
$a=-7, b=5$
and $d=-14$