A triangle is formed by the tangents at the point (2,2) on the curves y^2=2x and x^2+y^2=4…

Question

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 \_\_\_\_\_\_\_.

Answer: 10

Accepted Answer

Tangent to $y^2=2x$ at $(2,2)$: $y\cdot2 = 1(x+2) \implies T_1: x-2y+2=0$.

Tangent to $x^2+y^2=4x$ at $(2,2)$: $x(2)+y(2)=2(x+2) \implies T_2: y=2$.

Third line: $T_3: x+y+2=0$.

Vertices of the triangle formed by $T_1,T_2,T_3$:
- $T_1\cap T_2$: $x-4+2=0\implies x=2$, so $(2,2)$.
- $T_2\cap T_3$: $x+2+2=0\implies x=-4$, so $(-4,2)$.
- $T_1\cap T_3$: $x-2(-2-x)+2=0 \implies 3x+6=0\implies x=-2,\ y=0$, so $(-2,0)$.

Side lengths: $PQ=\sqrt{(2-(-4))^2+0}=\sqrt{20}$ wait, checking: distance $(2,2)$ to $(-4,2)$ is $6$; $(-4,2)$ to $(-2,0)$: $\sqrt{4+4}=\sqrt8=2\sqrt2$; $(-2,0)$ to $(2,2)$: $\sqrt{16+4}=\sqrt{20}=2\sqrt5$.

Using the source's labeled sides: $a=\sqrt{20},\ b=\sqrt8,\ c=6$, area $=\dfrac12\times6\times2=6$.

$$r = \frac{abc}{4\times\text{area}} = \frac{\sqrt{20}\cdot\sqrt8\cdot6}{24} = \frac{6\sqrt{160}}{24}=\sqrt{10}$$

$r^2 = 10$

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