ExamBro
ExamBro
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 \(...........\)

  1. A \(6\)
  2. B \(5\)
  3. C \(4\)
  4. D \(7\)
Verified Solution

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 \(x+y=2\) \(\Rightarrow 2 \sqrt{a^2-\left(\frac{a+a-2}{\sqrt{2}}\right)^2}=2\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app