A triangle is formed by the tangents at the p | vedclass

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(A) The curves are $S_1: y^2=2x$ and $S_2: x^2+y^2=4x$.The point $P(2,2)$ lies on both curves.Tangent $T_1$ to $S_1$ at $P(2,2)$ is given by $y(2) = x

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$10$

Vedclass · Answer

(A) The curves are $S_1: y^2=2x$ and $S_2: x^2+y^2=4x$.The point $P(2,2)$ lies on both curves.Tangent $T_1$ to $S_1$ at $P(2,2)$ is given by $y(2) = x+2$,which simplifies to $x-2y+2=0$.Tangent $T_2$ to $S_2$ at $P(2,2)$ is given by $x(2)+y(2) = 2(x+2)$,which simplifies to $2x+2y = 2x+4$,or $y=2$.The third line is $L_3: x+y+2=0$.To find the vertices of the triangle:Intersection of $T_1$ and $T_2$: $x-2(2)+2=0 \Rightarrow x=2$. So $P(2,2)$.Intersection of $T_1$ and $L_3$: $x-2y+2=0$ and $x+y+2=0$. Subtracting gives $3y=0 \Rightarrow y=0, x=-2$. So $Q(-2,0)$.Intersection of $T_2$ and $L_3$: $y=2$ and $x+y+2=0$ $\Rightarrow x+2+2=0$ $\Rightarrow x=-4$. So $R(-4,2)$.The vertices are $P(2,2)$,$Q(-2,0)$,and $R(-4,2)$.Side lengths:$PQ = \sqrt{(2 - (-2))^2 + (2-0)^2} = \sqrt{4^2 + 2^2} = \sqrt{16+4} = \sqrt{20}$.$QR = \sqrt{(-2 - (-4))^2 + (0-2)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8}$.$RP = \sqrt{(2 - (-4))^2 + (2-2)^2} = \sqrt{6^2 + 0^2} = 6$.Area of $\Delta PQR = \frac{1}{2} |x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)| = \frac{1}{2} |2(0-2) + (-2)(2-2) + (-4)(2-0)| = \frac{1}{2} |-4 + 0 - 8| = \frac{1}{2} |-12| = 6$.Circumradius $r = \frac{abc}{4\Delta} = \frac{\sqrt{20} \cdot \sqrt{8} \cdot 6}{4 \cdot 6} = \frac{\sqrt{160}}{4} = \frac{4\sqrt{10}}{4} = \sqrt{10}$.Therefore,$r^2 = 10$.

$A$ 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 $........$.

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