Prove

Prove that k1 and k2 are the equations of two circles. Find the equation of the line that passes through the centers of these circles.

k1: x²+y²+2x+4y+1=0
k2: x²+y²-8x+6y+9=0

Correct answer:

p : p=x+5y+11=0

Step-by-step explanation:

k_{1} : x^{2} + y^{2} + 2 x + 4 y + 1 = 0 k_{2} : x^{2} + y^{2} - 8 x + 6 y + 9 = 0 (x - x_{0})^{2} + (y - y_{0}) = r^{2} k_{1} : (x + 1)^{2} + (y + 2) = 2^{2} k_{2} : (x - 4)^{2} + (y + 3)^{2} = 4^{2} S_{1} [- 1, - 2] r_{1} = 2 S_{2} [4, - 3] r_{2} = 4 a_{x} + b_{y} + c = 0 a \cdot (- 1) + b \cdot (- 2) + c = 0 a \cdot 4 + b \cdot (- 3) + c = 0 a = 1 a + 2 b - c = 0 4 a - 3 b + c = 0 a = 1 P i v o t : R o w 1 \leftrightarrow R o w 2 4 a - 3 b + c = 0 a + 2 b - c = 0 a = 1 R o w 2 - \frac{1}{4} \cdot R o w 1 \to R o w 2 4 a - 3 b + c = 0 2.75 b - 1.25 c = 0 a = 1 R o w 3 - \frac{1}{4} \cdot R o w 1 \to R o w 3 4 a - 3 b + c = 0 2.75 b - 1.25 c = 0 0.75 b - 0.25 c = 1 R o w 3 - \frac{0 . 7 5}{2 . 7 5} \cdot R o w 2 \to R o w 3 4 a - 3 b + c = 0 2.75 b - 1.25 c = 0 0.091 c = 1 c = \frac{1}{0 . 0 9 0 9 0 9 0 9} = 11 b = \frac{0 + 1 . 2 5 c}{2 . 7 5} = \frac{0 + 1 . 2 5 \cdot 1 1}{2 . 7 5} = 5 a = \frac{0 + 3 b - c}{4} = \frac{0 + 3 \cdot 5 - 1 1}{4} = 1 a = 1 b = 5 c = 11 p : p = x + 5 y + 11 = 0

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