JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Two circles in the first quadrant of radii \(r_1\) and \(r_2\) touch the coordinate axes. Each of them cuts off an intercept of \(2\) units with the line \(x+y=2\). Then \(r_1^2+r_2^2-r_1 r_2\) is equal to \(...........\)
- A \(6\)
- B \(5\)
- C \(4\)
- D \(7\)
Answer & Solution
Correct Answer
(D) \(7\)
Step-by-step Solution
Detailed explanation
\(\text { Circle }(x-a)^2+(y-a)^2=a^2\) \(x^2+y^2-2 a x-2 a y+a^2=0\) \(\text { intercept }=2\) \(\Rightarrow 2 \sqrt{a^2-d^2}=2\) Where \(d=\) perpendicular distance of centre from line \(x+y=2\) \(\Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2\)…
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