Circle Ques 15
- The abscissae of the two points $A$ and $B$ are the roots of the equation $x^{2}+2 a x-b^{2}=0$ and their ordinates are the roots of the equation $y^{2}+2 p y-q^{2}=0$. Find the equation and the radius of the circle with $A B$ as diameter.
$(1984,4$ M)
Integer Answer Type Questions
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Solution:
Formula:
- Let $\left(x _1, y _1\right)$ and $\left(x _2, y _2\right)$ be the coordinates of points $A$ and $B$, respectively.
It is given that $x _1, x _2$ are the roots of $x^{2}+2 a x-b^{2}=0$ $\Rightarrow \quad x _1+x _2=-2 a$ and $x _1 x _2=-b^{2}$
Also, $y _1$ and $y _2$ are the roots of $y^{2}+2 p y-q^{2}=0$
$\Rightarrow \quad y _1+y _2=-2 p$ and $y _1 y _2=-q^{2}$
$\therefore$ The equation of circle with $A B$ as diameter is,
$\left(x-x _1\right)\left(x-x _2\right)+\left(y-y _1\right)\left(y-y _2\right)=0$
$\Rightarrow x^{2}+y^{2}-\left(x _1+x _2\right) x-\left(y _1+y _2\right) y+\left(x _1 x _2+y _1 y _2\right)=0$
$\Rightarrow x^{2}+y^{2}+2 a x+2 p y-\left(b^{2}+q^{2}\right)=0$
and radius $=\sqrt{a^{2}+p^{2}+b^{2}+q^{2}}$
BETA
