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 distinct points, then [2024]

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Solution

Q.
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 distinct points, then [2024]

1 5 < r < 9

2 3 < r < 7

3 0 < r < 7

4 $\frac{1}{2} < r < 7$

Ans.
(2)

Let $P : {(x + 1)}^{2} + {(y + 2)}^{2} = r^{2}$ and $Q : x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ be two given circles.

Q can be written as ${(x - 2)}^{2} + {(y - 2)}^{2} = 4$

Centre of circle P and Q are (–1, –2) and (2, 2) respectively

Distance between centre of circle is given by

$D = \sqrt{{(2 + 1)}^{2} + {(2 + 2)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 units$

For the intersection of circles, $D > | r_{P} - r_{Q} |$ and $D < (r_{P} + r_{Q})$ , where $r_{P}$ and $r_{Q}$ are radius of circle P and Q respectively

$\Rightarrow 5 > | r - 2 |$ and 5 < r + 2

$\Rightarrow - 5 < r - 2 < 5 \Rightarrow - 3 < r < 7$ ... (i)

and r > 3 ... (ii)

From (i) and (ii), 3 < r < 7.

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