Given-the-equation-of-two-circles-C-1-x-2-y-2-6x-4y-9-0-andC-2-x-2-y-2-2x-6y-9-0-find-the-equation-of-the-common-tangent-to-both-circles-

Question Number 110181 by Rio Michael last updated on 27/Aug/20

Given the equation of two circles

C_{1} : x^{2} + y^{2} - 6 x - 4 y + 9 = 0 and C_{2} : x^{2} + y^{2} - 2 x - 6 y + 9 = 0

find the equation of the common tangent to both circles .

Commented by bemath last updated on 27/Aug/20

Commented by bemath last updated on 27/Aug/20

c e n t r e p o i n t C_{1} (3, 2), r a d i u s = r_{1} = 2

c e n t r e p o i n t C_{2} (1, 3), r a d i u s = r_{2} = 1

s a y t h e c o m m o n t a n g e n t l i n e t o b o t h c i r c l e y = m x + n

o r m x - y + n = 0

(1) 2 = ∣ \frac{3 m - 2 + n}{\sqrt{13}} ∣\Rightarrow 2 \sqrt{13} =∣ 3 m + n - 2 ∣

3 m + n = 2 \pm 2 \sqrt{13} \dots (1)

(2) 1 =∣ \frac{m - 3 + n}{\sqrt{10}} ∣ \Rightarrow \sqrt{10} =∣ m + n - 3 ∣

m + n = 3 \pm \sqrt{10} \dots (2)

(1) - (2) \to 2 m = - 1 \pm 2 \sqrt{13} \mp \sqrt{10}

m = - \frac{1}{2} \pm \sqrt{13} \mp \frac{\sqrt{10}}{2}

n = \frac{7}{2} \pm \sqrt{10} \mp \sqrt{13} \pm \frac{\sqrt{10}}{2}

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