JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(C\) be the circle with centre at \((1, 1)\) and radius \(= 1\). If \(T\) is the circle centred at \((0, y),\) passing through origin and touching the circle \(C\) externally, then the radius of \(T\) is equal
- A \(\frac{1}{2}\)
- B \(\frac{1}{4}\)
- C \(\frac{{\sqrt 3 }}{{\sqrt 2 }}\)
- D \(\frac{5}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(C T^{2}=(1-0)^{2}+(1-y)^{2}\) also \(C T=1+y\) \(\therefore(1+y)^{2}=1+(1-y)^{2}\) \(1^{2}+2 y+y^{2}=1+1^{2}-2 y+y^{2}\) \(2 y+2 y=1+1^{2}+y^{2}-1^{2}-y^{2}\) \(4 y=1\) \(\Longrightarrow y=\frac{1}{4}\) Hence the radius of \(T\) is \(\frac{1}{4}\)
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