JEE Main 2022 · 28 Jun · Shift 1 · Mathematics Q30

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MathematicsInteger+4 / -1Question 30 of 90

Let the lines $y+2x=\sqrt{11}+7\sqrt7$ and $2y+x=2\sqrt{11}+6\sqrt7$ be normal to a circle $C:(x-h)^2+(y-k)^2=r^2$ . If the line $\sqrt{11}y-3x=\dfrac{5\sqrt{77}}3+11$ is tangent to the circle C, then the value of $(5h-8k)^2+5r^2$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 816

Centre is where the normals meet: $h=\frac{8\sqrt7}3$ , $k=\sqrt{11}+\frac{5\sqrt7}3$ . Then $5h-8k=-8\sqrt{11}$ , square 704.

Radius. Distance to the tangent: $\dfrac{|\sqrt{11}k-3h-\frac{5\sqrt{77}}3-11|}{\sqrt{20}}=\dfrac{8\sqrt7}{\sqrt{20}}$ , so $r^2=\frac{448}{20}$ and $5r^2=112$ .

Total $704+112=816$ .

Answer: 816

Topics:Tangents and normals Distance of a point from a line

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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 30