Question

$\text{A}$: If $x^2+y^2-2x+3y+k=0$ and $x^2+y^2+8x-6y-7=0$ cut each other orthogonally, then $k=10$
$\text{R}:$ The circles $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2+2g^{\prime }x+2f^{\prime }y+c^{\prime }=0$ cut each other orthogonally if $2g{g}^{\prime }+2f{f}^{\prime }=c+c^{\prime }$

• A. Both A and R are true and R is the correct explanation of A.
• B. Both A and R are true but R is not the correct explanation of A.
• C. A is true but R is false.
• D. A is false but R is true.

Answer 1

The correct option is D A is false but R is true.

$\text{A}:$

$S_1:x^2+y^2-2x+3y+k=0$
Radius, $r_1=\sqrt{\frac{13}{4}-k}$; centre, $C_1=(1,-\frac{3}{2})$
$S_2:x^2+y^2+8x-6y-7=0$
Radius, $r_2=\sqrt{25+7}=\sqrt{32}$; centre, $C_2=(-4,3)$

$\therefore$ By property of orthogonal circles, we can say
$r{{}_1}^2+r{{}_2}^2=(C_1{C}_2{)}^2$
$\Rightarrow \frac{13}{4}-k+32=25+\frac{81}{4}$
$\Rightarrow k=7-\frac{68}{4}$
$\Rightarrow k=7-16=-10$ $\to$ False
$\text{R}:$

$S_1:x^2+y^2+2gx+2fy+c=0$

Radius, $r_1=\sqrt{g^2+f^2-c}$; centre, $C_1=(g,f)$

$S_2:x^2+y^2+2g^{\prime }x+2f^{\prime }y+c^{\prime }=0$
Radius, $r_2=\sqrt{(g^{\prime }{)}^2+(f^{\prime }{)}^2-c^{\prime }}$; centre, $C_1=(g^{\prime },f^{\prime })$

If they orthogonal to each other then,
$r{{}_1}^2+r{{}_2}^2=(C_1{C}_2{)}^2$
$g^2+f^2-c+(g^{\prime }{)}^2+(f^{\prime }{)}^2-c^{\prime }=(g^{\prime }{)}^2+g^2-2g{g}^{\prime }+(f^{\prime }{)}^2+f^2-2f{f}^{\prime }$
$\Rightarrow c+c^{\prime }=2g{g}^{\prime }+2f{f}^{\prime }$ $\to$ True

Hence, option D.