Question

Assertion :Let $S_1:x^2+y^2+4x-6y+9=0$ and $S_2:x^2+y^2+6x-4y+11=0$ then $S_1$ bisect the circumference of $S_2$. Reason: The centre of $S_1$ lies on $S_2$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Equation of corrimon chord of $S_1$ & $S_2$ is
$(x^2+y^2+4x-6y+9)-(x^2+y^2+6x-4y+11)=0$
$\Rightarrow 2x+2y+2=0$ or $x+y+1=0$
Now centre of $S_{2}i.e.(-3,2)$ lies on it. Hence $x+y+1=0$ is the diameter of $S_2$
Hence $S_1$ bisect but circumference of circle $S_2$ & centre of $S_1$ i.e. $(-2,3)$ lies on $S_2=0$
Hence choice (a) is correct answer.
Ans: A