Question

The line $Ax+By+C=0$ cuts the the circle $x^2+y^2+ax+by+c=0$ in $P$ and $Q$. The line $A^{\prime }x+B^{\prime }y?+C^{\prime }=0$ cuts the circle $x^2+y^2+a^{\prime }x+b^{\prime }y+c^{\prime }=0$ in $R$ and $S$. If $P,Q,R,S$ are concyclic points, then

• A. $\left|\begin{array}{ccc}a+a^{\prime }& b+b^{\prime }& c+c^{\prime }\\ A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right|=0$
• B. $\left|\begin{array}{ccc}a-a^{\prime }& b-b^{\prime }& c-c^{\prime }\\ A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right|=0$
• C. $\left|\begin{array}{ccc}A(a+a^{\prime })& B(b+b^{\prime })& C(c+c^{\prime })\\ A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right|=0$
• D. None of these

Answer 1

The correct option is B $\left|\begin{array}{ccc}a-a^{\prime }& b-b^{\prime }& c-c^{\prime }\\ A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right|=0$

Equation of any circle through the points of intersection $P$ and $Q$ of line

$Ax+By+C=0$ ....(1)

and the circle $x^2+y^2+ax+by+c=0$ ...(2)

is $x^2+y^2+ax+by+c+\lambda (Ax+By+C)=0$

$\Rightarrow x^2+y^2+(a+\lambda A)x+(b+\lambda B)+c+\lambda C=0$ ...(3)

Again, equation of any circle through the point of intersection $R$ and $S$ of line

$A^{\prime }x+B^{\prime }y+C^{\prime }=0$ ....(4)

and circle $x^2+y^2+a^{\prime }x+b^{\prime }y+c^{\prime }=0$ ...(5)

is $x^2+y^2+a^{\prime }x+b^{\prime }y+c^{\prime }+\mu (A^{\prime }x+B^{\prime }y+C^{\prime })=0$

$\Rightarrow x^2+y^2+(a^{\prime }+\mu A^{\prime })x+(b^{\prime }+\mu B^{\prime })+c^{\prime }+\mu C^{\prime }=0$ ...(6)

If circle (3) and (6) are same, then points $P,Q,R,S$ will lie on the same circle ,i.e. points $p,Q,R,S$ will be concyclic.

Comparing the coefficients in (3) and (6), we get

$\frac{1}{1}=\frac{1}{1}=\frac{a+\lambda A}{a^{\prime }+\mu A^{\prime }}=\frac{b+\lambda A}{b^{\prime }+\mu B^{\prime }}=\frac{c+\lambda C}{c^{\prime }+\mu C^{\prime }}$

From these

$a-a^{\prime }+\lambda A-\mu A^{\prime }=0b-b^{\prime }+\lambda B-\mu B^{\prime }=0c-c^{\prime }+\lambda C-\mu C^{\prime }=0$

Eliminating $\lambda$ and $-\mu$ from above equation and writing the result in determinant from, we get

$\left|\begin{array}{ccc}a-a^{\prime }& A& A^{\prime }\\ b-b^{\prime }& B& B^{\prime }\\ c-c^{\prime }& C& C^{\prime }\end{array}\right|=\left|\begin{array}{ccc}a-a^{\prime }& b-b^{\prime }& c-c^{\prime }\\ A& B& C\\ A^{\prime }& B^{\prime }& C^{\prime }\end{array}\right|=0$