Question

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