Question

If $C$ is a circle described on the focal chord of the parabola $y^2=4x$ as diameter which is inclined at an angle of ${45}^{\circ }$ with the positive $x-$axis, then

• A. Radius of the circle is $2$ units
• B. The centre of circle is $(3,2)$
• C. The line $x+1=0$ touches the circle
• D. The circle $x^2+y^2+2x-6y+3=0$ is orthogonal to $C$

Answer 1

Answer: (B), (C)

The line $x+1=0$ touches the circle

Given parabola equation is $y^2=4x$
Let $P=(t_1^2,2t_1),Q=(t_1^2,2t_1)$
Let $PSQ$ be the focal chord, where $S(1,0)$ is the focus of the parabola
$t_1{t}_2=-1$
Since, circle described on focal chord as diameter always touches the directrix,
So, the line $x+1=0$ always touches the circle.
$\because$The slope of $PQ$ is
$\tan {45}^{\circ }=1\Rightarrow \frac{2t}{t^2-1}=1\Rightarrow t^2-2t-1=0$
Whose roots are $t_1,t_2$, so
$t_1+t_2=2,t_1{t}_2=-1$

The length of focal chord
$=4a{\text{cosec}}^2\theta =4\times 1\times (\sqrt{2}{)}^2=8$
So, the radius of circle is
$r=\frac{8}{2}=4\text{units}$

Now, the equation of the circle described on focal chord as diameter is
$(x-t_1^2)(x-t_2^2)+(y-2t_1)(y-2t_2)=0\Rightarrow x^2-(t_1^2+t_2^2)x+t_1^2{t}_2^2+y^2-(2t_1+2t_2)y+4t_1{t}_2=0\Rightarrow x^2+y^2-(t_1^2+t_2^2)x-2(t_1+t_2)y-3=0\Rightarrow x^2+y^2-\left[\right(t_1+t_2{)}^2-2t_1{t}_2]x-4y-3=0\Rightarrow x^2+y^2-6x-4y-3=0$

Centre $\equiv (3,2)$

Circles $x^2+y^2-6x-4y-3=0$ and $x^2+y^2+2x-6y+3=0$ are not orthogonal
$\because 2(g{g}^{\prime }+f{f}^{\prime })\ne c+c^{\prime }$