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Number of Common Tangents to Two Circles in Different Conditions

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Question

The number of common tangents to the circles $x^{2} + y^{2} - 4 x - 2 y + 1 = 0$ and $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$ is

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Solution

Given circles are
$x^{2} + y^{2} - 4 x - 2 y + 1 = 0$ and $x^{2} + y^{2} - 6 x - 4 y + 4 = 0$
The centre and radius are,
$C_{1} (2, 1), r_{1} = 2 C_{2} (3, 2), r_{2} = 3 C_{1} C_{2} = \sqrt{(3 - 2)^{2} + (2 - 1)^{2}} = \sqrt{2} \Rightarrow | r_{1} - r_{2} | < C_{1} C_{2} < r_{1} + r_{2}$

Hence, the number of common tangents is $2.$

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Q. The number of common tangents to the circles

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

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

Q. I : The equations to the direct common tangents to the circles

x^{2} + y^{2} + 6 x + 4 y + 4 = 0

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

y - 1 = 0

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II : The equations to the transverse common tangents to the circles

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

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

x - 1 = 0, 3 x + 4 y - 21 = 0

Q. The internal common tangents of the circles

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

x^{2} + y^{2} + 6 x + 6 y + 9 = 0

Q. The length of the transverse common tangent of the circles

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

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

Q. The number of common tangents to the circles

x^{2} + y^{2} + 6 x + 6 y + 14 = 0

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

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Number of Common Tangents to Two Circles in Different Conditions

Standard XIII Mathematics

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