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# Find the equation of the circle passing through the points $(2,3)$ and $(–1,1)$ and whose centre is on the line $x–3y–11 = 0$.

$\begin {array} {1 1} (A)\;x^2+y^2+7x-5y+14=0 & \quad (B)\;x^2+y^2-7x-5y+14=0 \\ (C)\;x^2+y^2-7x-5y-14=0 & \quad (D)\;x^2+y^2-7x-5y+14=0 \end {array}$

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Toolbox:
• Equation of a circle with centre (h,k) and radius r is given as $(x-h)^2+(y-k)^2=r^2$
Step 1 :
Let the equation of the required circle be $( x-h)^2+(y-k)^2=r^2$-----(1)
It is given that the circle passes through the point (2,3) and (-1, 1)
$\therefore (2-h)^2 +(3-k)^2=r^2$ and
$(-1-h)^2+(1-k)^2=r^2$
$\Rightarrow (2-h)^2+(3-k)^2=(-1-h)^2+(1-k)^2$
On expanding
$4-4h+h^2+9-6k+k^2=1+2h+h^2+1-2k+k^2$
$\Rightarrow 4-4h+9-6k=1+2h+1-2k$
$\Rightarrow 6h+4k=11$------(2)
It is also given that the centre(h, k) of the circe lies on the line $x-3y-11=0$.
$\therefore h - 3k=11$---------(3)
Step 2 :
Solving equation (2) and (3) we get,
$h-3k=11 ( \times 6)$
$6h+4k=11$
_____________
$6h-18k=66$
$6h+4k=11$
$(-) \: \: (-) \: \: (-)$
_______________
$\quad -22k=55$
$\Rightarrow k = -\large\frac{5}{2}$
Substituting for k in equation (3) we get,
$h = 3 \bigg( -\large\frac{5}{2} \bigg) $$=11 \Rightarrow h = 11-\large\frac{15}{2} (i.e) h = \large\frac{7}{2} Hence the coordinates of the centre are \bigg( \large\frac{7}{2}$$, -\large\frac{5}{2} \bigg)$
Substituting for h and k in equation (1) we get,
$\bigg( 2 - \large\frac{7}{2} \bigg)^2+ \bigg( 3+ \large\frac{5}{2} \bigg)^2=r^2$
(i.e) $\bigg( \large\frac{4-7}{2} \bigg)^2$$+ \bigg( \large\frac{6+5}{2} \bigg)^2$$=r^2$
(i.e) $\large\frac{9}{4} $$+ \large\frac{121}{4}$$=r^2$
(i.e) $r^2=\large\frac{130}{4} \Rightarrow r = \large\frac{\sqrt{130}}{2}$
Hence the equation of the circle is
$\bigg( x - \large\frac{7}{2} \bigg)^2$$+ \bigg( y+ \large\frac{5}{2} \bigg)^2$$=\large\frac{130}{4}$
On expanding we get,
$4x^2-28x+49+4y^2+20y+25=130$
$\Rightarrow 4x^2+4y^2-28x+20y-56=0$
dividing throughout by 4 we get,
$x^2+y^2-7x+5y-14=0$
This is the equation of the required circle.
answered Jul 13, 2014