JEE Main 2022 · 29 Jul · Shift 2 · Mathematics Q28

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MathematicsInteger+4 / -1Question 28 of 90

Let AB be a chord of length 12 of the circle $(x-2)^2+(y+1)^2=\dfrac{169}4$ . If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P from chord AB is equal to _______.

Your answer

Answer and worked solution

Official answer: 72

Distances from the centre. $r=\frac{13}2$ , half chord 6, so the chord is $d=\frac52$ away. P is at $\frac{r^2}d=\frac{169}{10}$ .

P to AB $=\frac{169}{10}-\frac52=\frac{72}5$ ; five times is 72.

Answer: 72

Topics:Chord of contact Tangents and normals

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JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 28