Tangents are drawn to the parabola $y^2=4x$ from the point P (6,5) to the touch the parabola at Q and R.

Question

$C_1$ is the circle which touches the parabola at Q and $C_2$ is the circle which touches the parabola at R. Both circles pass through the focus of the parabola. Find the radius of circle $C_2$

The equation of tangent to the parabola $$y=mx+\frac am$$ $$5=6m+\frac 1m$$ $$m=\frac 12 , \frac 13$$

Therefore, equation of tangents will be $$x-2y+4=0$$ and $$x-3y+9$$

The point of intersections with the parabola $y^2=4x$ were found out to be $(4,4)$ and $(9,6)$

Let R be $(9,6)$. Hence circle $C_2$ passes through (9,6) and focus (1,0)

This data isn’t enough to find the radius of the circle. How do I get more information?

Answer 1

Hint:

If the equation of $C_2$ is $$(x-h)^2+(y-k)^2=r^2$$

$$(1-h)^2+(0-k)^2=r^2=(9-h)^2+(6-k)^2\implies8(2h-10)+6(2k-6)\ \ \ \ (1)$$

Again, the gradient of $C_2$ at$(9,6)$

$$=-\dfrac{9-h}{6-k}$$ which should be $=$ the gradient of the parabola at $(9,6)$ $$\dfrac{4}{2\cdot6}$$

Solve for $h,k$

Answer 2

Find the equation of both the circles using $S+kL=0$.

$L$ is the equation of the tangent to the parabola at points of contact. Point of contact may be considered as point circles.

Since point of contact comes out to be $(4,4)$ and $(9,6)$ and the equation of tangents are: $2y=x+4$ and $3y=x+9$

Circle through $(9,6)$ is $(x-9)^2+(y-6)^2+k(3y-x-9)=0$

Since it passes through the focus $(1,0)$, therefore, putting the values of $X$ and $Y$ we get $K=10$.

Hence the equation comes out to be: $(x-9)^2+(y-6)^2+10(3y-x-9)=0$ or $x^2+y^2-28x+18y+108=0$

Similarly, for the second circle through $(4,4)$: $(x-4)^2+(y-4)^2+k(2y-x-4)=0$

Since it passes through $(1,0)$ we get $K$ as $5$. We have got both the circles and we can find the radius.

Using this method we get the centre of circles as $(13/2,-1)$ and $(14,-9)$.

Therefore using the centres and the point $(1,0)$ we can get the radius using the distance formula.

$r_1$ comes out to be $\sqrt{125/4}$ and $r_2$ comes out to be $\sqrt{250}$.

Answer 3

By homogenization, $$Yy - 2X - 2x=0 \tag{1}$$ is the equation for tangent of parabola at $(x,y)$, plug $(6,5)=(X,Y)$ into this:

$$ 5y -2x=12 \tag{2}$$

The above is the equation of the chord of contact, we now find the intersection of the above equation with the parabola $y^2-4x=0$, by manipulation, we find an equation in $y$:

$$(y-6)(y-4) = 0 \tag{3}$$

From the above, we find the point where the chord cuts the parabola are $(x,y) = \{ (9,6) , (4,4)\}$, plugging these points into $(1)$:

$$ 3Y-X-9=0 ; (x,y)=(9,6)\tag{4}$$ $$ 2Y- X-4=0 ; (x,y) = (4,4)\tag{5}$$

By Family of circles tangent to a line at each point:

$$ (x-9)^2 +(y-6)^2 + \lambda( 3Y-X-9) =0; (x,y)=(9,6)$$

$$ (x-4)^2 + (y-4)^2 + \lambda' (2Y-X-4)=0; (x,y) = (4,4)$$

Upon substitution of the focal point $(1,0)$ we find $\lambda= 10$ and $\lambda'=5$, hence the radius can be calculated by standard result as $\frac{5 \sqrt{5}}{2}$ and $ 5 \sqrt{10}$