Find all pairs of real numbers (x,y) such that x2−2x+y2+4y+5+x2−8x+y2−4y+20=516x2+9y2=68.
Solution
Consider orthogonal coordinate system and the points A(1,−2) and B(4,2). Since x2−2x+y2+4y+5x2−8x+y2−4y+20=(x−1)2+(y+2)2=(x−4)2+(y−2)2, we conclude that the solutions of the first equation are all points C(x,y) such that ∣CA∣+∣CB∣=5. But ∣AB∣=(4−1)2+(2−(−2))2=16+9=5=∣CA∣+∣CB∣, and thus the point C lies on the line segment AB. The equation of the line AB is 3y=4x−10 so we want to find the solutions of the two equations satisfying x∈[1,4]. Now the solution of the system is given by 32x2−80x+100=68⇔2x2−5x+2=0⇔x1=21,x2=2. Only x2 belongs to the interval [1,4], thus the only solution is (2,−32).
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Source: MathNet,
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