Animated Solution for Mathematics - Conic Sections: A triangle is formed by the tangents at the point $(2,2)$ on the curves $y^2=2x$ and $x^2+y^2=4x$, and the line $x+y+2=0$. If r is the radius of its circumcircle, then $r^2$ is equal to

We are given a parabola $y^2=2x$ and a circle $x^2+y^2=4x$, both intersecting at the point $P(2,2)$. Our objective is to determine the circumradius of the triangle formed by the tangents to these curves at $P$ and the line $x+y+2=0$.

To find the tangents, we invoke the $T=0$ method. For any conic, the tangent at $(x_1,y_1)$ is found by replacing $x^2$ with $x{x}_1$, $y^2$ with $y{y}_1$, $x$ with $\frac{x+x_1}{2}$, and $y$ with $\frac{y+y_1}{2}$.

For the parabola $y^2=2x$, the tangent at $P(2,2)$ is: This simplifies to $2y=x+2$, or $L_1:x-2y+2=0$.

For the circle $x^2+y^2-4x=0$, we apply the same transformation: Expanding this, we get $2x+2y-2x-4=0$. The $x$ terms cancel, leaving $2y=4$, or $L_2:y=2$.

We now have three lines defining our triangle: $L_1:x-2y+2=0$ $L_2:y=2$ $L_3:x+y+2=0$

To find the vertices, we solve for their intersections: 1. Intersection of $L_2$ and $L_3$: Substituting $y=2$ into $x+y+2=0$ gives $x+4=0$, so $x=-4$. Thus, vertex $Q$ is $(-4,2)$. 2. Intersection of $L_1$ and $L_3$: Subtracting the equations $(x-2y+2)-(x+y+2)=0$ yields $-3y=0$, so $y=0$. Substituting $y=0$ into $L_3$ gives $x=-2$. Thus, vertex $R$ is $(-2,0)$. 3. The third vertex is the point of tangency $P(2,2)$.

The vertices of the triangle are $P(2,2)$, $Q(-4,2)$, and $R(-2,0)$. We calculate the side lengths: $c=PQ=\sqrt{(2-(-4){)}^2+(2-2{)}^2}=6$ $a=QR=\sqrt{(-4-(-2){)}^2+(2-0{)}^2}=\sqrt{(-2{)}^2+2^2}=\sqrt{8}$ $b=RP=\sqrt{(2-(-2){)}^2+(2-0{)}^2}=\sqrt{4^2+2^2}=\sqrt{20}$

The area $\Delta$ of the triangle is:

Finally, we use the circumradius formula $R=\frac{abc}{4\Delta }$:

Squaring the circumradius, we obtain:

The final value of the circumradius squared is 10.