If the circles ( x + 1 ) 2 + ( y + 2 ) 2 = r 2 and x 2 + y 2 - 4 x - 4 y + 4 = 0 intersect at exactly two…

If the circles (x+1)2+(y+2)2=r2 and x2+y2-4x-4y+4=0 intersect at exactly two distinct points, then
  1. 5<r<9
  2. 0<r<7
  3. 3<r<7
  4. 12<r<7

Solution

Given, $S_1 \equiv (x+1)^2 + (y+2)^2 = r^2$ and $S_2 \equiv x^2 + y^2 - 4x - 4y + 4 = 0$ or $S_2 \equiv (x-2)^2 + (y-2)^2 = 2^2$ Now, if two circles intersect at two distinct points then $|r_1 - r_2| < C_1 C_2 < r_1 + r_2$. $\Rightarrow |r - 2| < \sqrt{3^2 + 4^2} < r + 2$ $\Rightarrow |r - 2| < 5 < r + 2$ $\Rightarrow |r - 2| < 5$ and $r + 2 > 5$ $\Rightarrow -5 < r - 2 < 5$ and $r > 3$ $\Rightarrow -3 < r < 7$ and $r > 3$ $\Rightarrow 3 < r < 7$

Asked in: JEE Main 2024 (30 Jan Shift 1)

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