JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let a circle \(\mathrm{C}\) of radius \(1\) and closer to the origin be such that the lines passing through the point \((3,2)\) and parallel to the coordinate axes touch it. Then the shortest distance of the circle \(\mathrm{C}\) from the point \((5,5)\) is :
- A \(2 \sqrt{2}\)
- B \(5\)
- C \(4 \sqrt{2}\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
Coordinates of the centre will be \((2,1)\) Equation of circle will be \( (\mathrm{x}-2)^2+(\mathrm{y}-1)^2=1 \) \( \mathrm{QC}=\sqrt{(5-2)^2+(5-1)^2} \) \( \mathrm{QC}=5\) shortest distance \( =\mathrm{RQ}=\mathrm{CQ}-\mathrm{CR} \) \( =5-1 \) \( =4\)
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