If the two circles $(x - 1)^{2} + (y - 3)^{2} = r^{2}$ and $x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct point, then -

r > 2

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2 < r < 8

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r < 2

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r = 2

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Solution

The correct option is A $r > 2$
$∵$ Two circles intersect at 2 distinct points, hence:
$C_{1} C_{2} < (r_{1} + r_{2}) D i s t a n c e b e t w e e n c e n t r e s \Rightarrow \sqrt{(4 - 1)^{2} + (- 1 - 3)^{2}} < r + \sqrt{4^{2} + 1^{2} - 8} \Rightarrow \sqrt{9 + 16} < r + 3. \Rightarrow \sqrt{25} < r + 3. \Rightarrow r + 3 > 5 \Rightarrow r + 3 > 5 \Rightarrow r > 2.$

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Similar questions

(x - 1)^{2} + (y - 3)^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

(x - 1)^{2} + (y - 3)^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

If the cicles $(x - 1)^{2} + (y - 3)^{2} = r^{2} a n d x^{2} + y^{2} - 8 x + 2 y + 8 = 0$ intersect in two distinct points, then

{(x - 1)}^{2} + {(y - 3)}^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

{(x - 1)}^{2} + {(y - 3)}^{2} = r^{2}

x^{2} + y^{2} - 8 x + 2 y + 8 = 0

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