Medium MCQ +4 / -1 PYQ · JEE Mains 2022

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 _____________.

  1. A 12 Correct answer
  2. B 13
  3. C 14
  4. D 16

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

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