JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(r_{1}\) and \(r_{2}\) be the radii of the largest and smallest circles, respectively, which pass through the point \((-4,1)\) and having their centres on the circumference of the circle \(x^{2}+y^{2}+2 x+4 y-4= 0.\) If \(\frac{r_{1}}{r_{2}}=a+b \sqrt{2}\), then \(a+b\) is equal to:
- A \(3\)
- B \(11\)
- C \(5\)
- D \(7\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
Centre of smallest circle is \(\mathrm{A}\) Centre of largest circle is \(\mathrm{B}\) \(r_{2}=|C P-C A|=3 \sqrt{2}-3\) \(r_{1}=C P+C B=3 \sqrt{2}+3\) \(\frac{r_{1}}{r_{2}}=\frac{(3 \sqrt{2}+3)^{2}}{9}=(\sqrt{2}+1)^{2}=3+2 \sqrt{2}\) \(a=3, b=2\)
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