Definition of a circle

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Question

Equation of a circle with centre $(4, 3)$ touching the circle
$x^{2} + y^{2} = 1$ , is

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Solution

Let the circle be $x^{2} + y^{2} - 8 x - 6 y + k = 0$ which touches the circle $x^{2} + y^{2} = 1 .$ The equation of the common tangent to these circles is given by
$S_{1} - S_{2} = 0 \Rightarrow 8 x + 6 y - 1 - k = 0$
This is a tangent to the circle $x^{2} + y^{2} = 1$
$∴ \pm 1 = \frac{k + 1}{\sqrt{8^{2} + 6^{2}}} \Rightarrow k^{2} + 1 = \pm 10 \Rightarrow k = - 11$ or, $9$
Hence, the equations of the circles are
$x^{2} + y^{2} - 8 x - 6 y + 9 = 0$ and, $x^{2} + y^{2} - 8 x - 6 y - 11 = 0$

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Definition of a circle

Remember concepts with our Masterclasses.

Equation of a circle with centre (4, 3) touching the circlex2+y2=1, is

Ready to Test Your Skills?

free study material

Equation of a circle with centre $(4, 3)$ touching the circle
$x^{2} + y^{2} = 1$ , is