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Length of Common Chord

For the given...

Question

For the given circles $x^{2} + y^{2} + 2 x + 4 y - 20 = 0$ and $x^{2} + y^{2} + 6 x - 8 y + 10 = 0$ , which of the following is/are correct?

The number of common tangents is

2.

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The number of common tangents is

3.

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Length of the common tangent is

(1500)^{1 / 4}

units

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Length of the common tangent is

(1000)^{1 / 4}

units

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Solution

The correct option is C Length of the common tangent is $(1500)^{1 / 4}$ units
Given circles are
$x^{2} + y^{2} + 2 x + 4 y - 20 = 0$ and $x^{2} + y^{2} + 6 x - 8 y + 10 = 0$
The centre and radius are,
$C_{1} = (- 1, - 2), r_{1} = \sqrt{1 + 4 + 20} = 5 C_{2} = (- 3, 4), r_{2} = \sqrt{9 + 16 - 10} = \sqrt{15}$
Now,
$C_{1} C_{2} = \sqrt{2^{2} + 6^{2}} = \sqrt{40} \Rightarrow r_{1} + r_{2} > C_{1} C_{2} > | r_{1} - r_{2} |$
Therefore, the circles intersect in two distinct points, so they have $2$ common tangents.

Now, the length of common tangent (DCT)
$= \sqrt{(d)^{2} - (r_{1} - r_{2})^{2}} = \sqrt{40 - (5 - \sqrt{15})^{2}} (∵ d = C_{1} C_{2}) = (1500)^{1 / 4}$

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Length of Common Chord

Standard XII Mathematics

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For the given circles x2+y2+2x+4y−20=0 and x2+y2+6x−8y+10=0, which of the following is/are correct?

For the given circles $x^{2} + y^{2} + 2 x + 4 y - 20 = 0$ and $x^{2} + y^{2} + 6 x - 8 y + 10 = 0$ , which of the following is/are correct?