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Number of Common Tangents to Two Circles in Different Conditions

The circles x...

Question

The circles $x^{2} + y^{2} - 2 x - 4 y = 0$ and $x^{2} + y^{2} - 8 y - 4 = 0$ have

2

common tangents.

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3

common tangents.

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4

common tangents.

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1

common tangent.

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Solution

The correct option is D $1$ common tangent.
Given circle
$x^{2} + y^{2} - 2 x - 4 y = 0$ and $x^{2} + y^{2} - 8 y - 4 = 0$
The centre and radius of the circles are
$C_{1} = (1, 2), r_{1} = \sqrt{5} C_{2} = (0, 4), r_{2} = 2 \sqrt{5}$
Now,
$C_{1} C_{2} = \sqrt{1^{2} + 2^{2}} = \sqrt{5} r_{2} - r_{1} = \sqrt{5} C_{1} C_{2} = r_{2} - r_{1}$
Therefore, they touch internally, so $1$ common tangent.

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Similar questions

Circles $x^{2} + y^{2} - 2 x - 4 y = 0 a n d x^{2} + y^{2} - 8 y - 4 = 0$

Examine if the two circles $x^{2} + y^{2} - 2 x - 4 y = 0 a n d x^{2} + y^{2} - 8 y - 4 = 0$ _________.

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

x^{2} + y^{2} + 4 x + 4 y - 1 = 0

x^{2} + y^{2} - 2 x - 4 y = 0

x^{2} + y^{2} - 8 y - 4 = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

x^{2} + y^{2} - 4 = 0, x^{2} + y^{2} - 6 x - 8 y + 10 = 0

x^{2} + y^{2} + 2 x - 4 y - 2 = 0

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