Definition of a circle

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Question

Square of the length of the greatest common tangent to the circles given by $x^{2} + y^{2} - 4 = 0 and x^{2} + y^{2} - 12 x + 27 = 0$ is ____.

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Solution

For circle, $k^{2} + y^{2} - 4 = 0$ , centre is $C_{1} (0, 0)$ and radius is $r_{1} = 2$ .
For circle, $x^{2} + y^{2} - 12 x + 27 = 0$ , centre is $C_{2} (6, 0)$ and radius is $r_{2} = 3 .$
Length of greatest common tangent
$\begin{array}{l} = \sqrt{{(C_{1} C_{2})}^{2} - {(r_{1} - r_{2})}^{2}} \\ = \sqrt{36 - 1} = \sqrt{35} \end{array}$

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Definition of a circle

Remember concepts with our Masterclasses.

Square of the length of the greatest common tangent to the circles given by x2+y2−4=0 and x2+y2−12x+27=0 is ____.

For circle,k2+y2−4=0, centre is C1(0,0)and radius is r1=2.For circle, x2+y2−12x+27=0, centre is C2(6,0)and radius is r2=3.Length of greatest common tangent =C1C22−r1−r22=36−1=35

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Square of the length of the greatest common tangent to the circles given by $x^{2} + y^{2} - 4 = 0 and x^{2} + y^{2} - 12 x + 27 = 0$ is ____.