Let the lines y+2x=√11+7√7 and 2y+x=2√11+6√7 be normal to a circle C:(x-h)^2+(y-k)^2= ^2. If the line √11y-3x= 5√77 3+11 is tangent to the circle C,…
JEE Main 2022 — Mathematics Coordinate Geometry
2022integerhard
Let the lines y+2x=11+77 and 2y+x=211+67 be normal to a circle C:(x−h)2+(y−k)2=r2. If the line 11y−3x=3577+11 is tangent to the circle C, then the value of (5h−8k)2+5r2 is equal to ______.
Official previous-year question
Held on 28 Jun 2022 · Verified 6 Jul 2026.
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Solution
Equations of normal are
y+2x=11+77...(i)
2y+x=211+67...(ii)
Now the center of the circle is point of intersection of the normals i.e. solving (i)&(ii), we get the point of intersection as
(387,11+357)≡(h,k)
The equation of tangent is 11y−3x=3577+11
The radius will be perpendicular distance of tangent from center
i.e. r=11+9∣11387−3(11+357)−3577−11∣=457
Hence (5h−8k)2+5r2=816
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