If the circles x^2+y^2-16x-20y+164=r^2 and (x-4)^2+(y-7)^2=36 intersect at two distinct points, then:

Question

Accepted Answer

Correct answer: D. (d) JEE Main 2019 Consider the equation of circles as, x^2+y^2-16x-20y+164=r^2 i.e., (x-8)^2+(y-10)^2=r^2 … (i) and (x-4)^2+(y-7)^2=36 … (ii) Both the circles intersect each other at two distinct points. Distance between centres = (8-4)^2+(10-7)^2=5 ∴(r-6)<5<(r+6) ∴ If (r-6)<5 ⇒rϵ (1 ,11 ) … (iii) And (r+6)>5 ⇒rϵ(- ,-11) ∪(-1 , ) … (iv) From (iii) & (iv), rϵ (1 ,11 )

$If the circles x^{2} + y^{2} - 16 x - 20 y + 164 = r^{2} and (x - 4)^{2} + (y - 7)^{2} = 36 intersect at two distinct points, then:$

Solution

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