Question

If the lines $x-2y+3=0$ and $3x+ky+7=0$ cut the coordinate axes in concyclic points, then

• A. $k=\frac{3}{2}$
• B. $k=-\frac{3}{2}$
• C. Equation of circle passing through those points is ${(x+\frac{8}{3})}^2+{(y-\frac{37}{12})}^2=\frac{1385}{12}$
• D. Center of circle is $(-\frac{8}{3},\frac{37}{12})$

Answer 1

Answer: (B), (D)

Center of circle is $(-\frac{8}{3},\frac{37}{12})$

If lines $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ cut axes at concyclic points then
(i) $a_1{a}_2=b_1{b}_2$
(ii) Circle equation is $(a_{1}x+b_{1}y+c_1)(a_{2}x+b_{2}y+c_2)-(a_1{b}_2+a_2{b}_1)xy=0$
So, for
$x-2y+3=0$ and $3x+ky+7=0$
$1\left(3\right)=-2\left(k\right)$
$\Rightarrow k=-\frac{3}{2}$
circle equation passing through concyclic point is $(x-2y+3)(6x-3y+14)-(-3-12)xy=0$
$6x^2+6y^2+32x-37y+42=0$
$\Rightarrow x^2+y^2+\frac{16}{3}x-\frac{37}{6}y+7=0$
Centre$=(\frac{-8}{3},\frac{37}{12})$
radius$=\sqrt{{\left(\frac{8}{3}\right)}^2+{\left(\frac{37}{12}\right)}^2-7}$
$=\frac{1}{3}\sqrt{64+\frac{1369}{16}-63}$
$=\frac{\sqrt{1385}}{12}$

Alternate Solution:
For points to be concyclic all $4$ points should be in $2^{\text{nd}}$ Quadrant
Given $AB:x-2y+3=0$
and $2p=3/2-7/k$
$\because$ center will lie on perpendicular bisector of $AB$ whose equation will be
$2x+y=-9/4\cdots \left(1\right)$
Now putting the center coordinates in $\left(1\right)$
$-\frac{16}{3}+\frac{3}{4}-\frac{7}{2k}=-\frac{9}{4}\Rightarrow k=-3/2$
Hence center coordinates $(-\frac{8}{3},\frac{37}{12})$
radius $=\sqrt{{(-\frac{8}{3}+3)}^2+{\left(\frac{37}{12}\right)}^2}$
$=\frac{\sqrt{1385}}{12}$ units
Hence, circle equation is ${(x+\frac{8}{3})}^2+{(y-\frac{37}{12})}^2=\frac{1385}{144}$