Two points And A Circle Of Life!

For shortest path, we have to go straight line to circle and then straight line to $(0,100)$

We know that a curve who sum of distances from two points is constant is an ellipse

Let's build ellipses with $P(-100,100)$ and $Q(0,100)$ as foci till it touches the circle as at that point, sum of distances is minimum.

Now, if ellipse touches circle at point $R$ , then ray $PR$ after reflection from ellipse will pass from $Q$ by reflexion property of ellipse. Now, as circle and ellipse touch each other, at contact point both have equal slope, ray $PR$ after reflection from circle too will pass from $Q$ .

So, we have to find out a point $R$ on circle such ray $PR$ after reflection from circle will pass through $Q$ .

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Let point $R$ be $(50\cos\theta, 50\sin\theta)$

Now, equation of tangent at $R$ will be $x\cos\theta + y\sin\theta = 50$

Now, as $PR$ after reflection from circle or tangent will pass from $Q$ , mirror image of $P$ about tangent is collinear with $QR$ . Let mirror image be $P'(a,b)$

For mirror image of $P$ about tangent,

$\frac{a+100}{\cos\theta} = \frac{b-100}{\sin\theta} = -2\frac{a\cos\theta + b\sin\theta -50}{\cos^2\theta + \sin^2\theta}$

$(a,b) = (200\cos^2\theta - 200\sin\theta\cos\theta + 100\cos\theta - 100,-200\sin^2\theta + 200\sin\theta\cos\theta + 100\sin\theta + 100)$

Now, as $P', Q, R$ are collinear

$\frac{b- 50\sin\theta}{a-50\cos\theta} = \frac{50\sin\theta - 100}{50\cos\theta - 0}$

$\sin\theta + 4\cos^2\theta - 4\sin\theta\cos\theta + 2\cos\theta - 2 = 0$

$32\cos^4\theta + 8\cos^3\theta - 27\cos^2\theta + 3 = 0$

$\cos\theta = -1, -0.3395, 0.4, 0.69$

Now, we know from diagram and common sense that ellipse touches circle in $2^{nd}$ quadrant and that too not too down. So, $\cos\theta = -0.3395$ is only logical value. So, $\sin\theta = 0.9406$

So, point $R = (-16.975, 47.03)$ .

So, length of path $L = PR + RQ$

$L = 98.483 + 55.623$

$\color{#3D99F6}{\boxed{L = 154.09}}$