Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^{2}-4 x-6=0$ and the ordinates of $\mathrm{P}$ and $\mathrm{Q}$ be the roots of $y^{2}+2 y-7=0$. If $\mathrm{PQ}$ is a diameter of the circle $x^{2}+y^{2}+2 a x+2 b y+c=0$, then the value of $(a+b-c)$ is _____________.
Solution
<p>Abscissae of PQ are roots of ${x^2} - 4x - 6 = 0$</p>
<p>Ordinates of PQ are roots of ${y^2} + 2y - 7 = 0$</p>
<p>and PQ is diameter</p>
<p>$\Rightarrow$ Equation of circle is</p>
<p>${x^2} + {y^2} - 4x + 2y - 13 = 0$</p>
<p>But, given ${x^2} + {y^2} + 2ax + 2by + c = 0$</p>
<p>By comparison $a = - 2,b = 1,c = - 13$</p>
<p>$\Rightarrow a + b - c = - 2 + 1 + 13 = 12$</p>
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
This question is part of PrepWiser's free JEE Main question bank. 40 more solved questions on Circles are available — start with the harder ones if your accuracy is >70%.