Question

If the equation of the parabola whose focus is the point of intersection of $x+y=3,$ $x-y=1$ and directrix is $x-y+5=0,$ is $a{x}^2+bxy+c{y}^2+dx+ey-15=0,$ then the radius of the circle $a{x}^2+c{y}^2+dx+ey-10=0$ is equal to

• A. $10$
• B. $5$
• C. $12$
• D. $13$

Answer 1

The correct option is A $10$

$x+y=3\cdots \left(1\right)$
$x-y=1\cdots \left(2\right)$
The point of intersection of $\left(1\right)$ and $\left(2\right)$ is $(2,1)$
Directrix $:x-y+5=0$
Equation of parabola is $\sqrt{(x-2{)}^2+(y-1{)}^2}=\frac{|x-y+5|}{\sqrt{2}}$
$\Rightarrow x^2+2xy+y^2-18x+6y-15=0$
$\therefore a=1,b=2,c=1,d=-18,e=6$

Now, equation of the circle is $x^2+y^2-18x+6y-10=0$
$\therefore$ Radius $=\sqrt{(-9{)}^2+(3{)}^2+10}=10$