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

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Solution

Two circles with radius $r_{1}, r_{2}$ touch
each other when distance between
centers $r_{1} + r_{2}$
Given circles center $x^{2} + y^{2} = 1$ (0,0)
$r_{1} + r_{2} = \frac{(4, 3)}{\sqrt{(4 - 0)^{2} + (3 - 0)^{2}}} = 5$
$\Rightarrow r_{2} = 4.$
circle centered at (4,3) and has radius 4
$(x - 4)^{2} + (y - 3)^{2} = 16$

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