A curve is traced by a point \( P(x, y) \) which moves such that its distance from the point \( A(1, -2) \) is twice its distance from the point \( B(4, -3) \). Determine the equation of the curve.

∣ P A ∣ = ( 1 − x ) 2 + ( − 2 − y ) 2 ​ ∣ PB ∣ = ( 4 − x ) 2 + ( − 3 − y ) 2 ​ ∣ P A ∣ = 2∣ PB ∣ ( 1 − x ) 2 + ( − 2 − y ) 2 ​ = 2 ( 4 − x ) 2 + ( − 3 − y ) 2 ​ ( 1 − x ) 2 + ( − 2 − y ) 2 = 4 [( 4 − x ) 2 + ( − 3 − y ) 2 ] 1 − 2 x + x 2 + 4 + 4 y + y 2 = 4 ( 16 − 8 x + x 2 + 9 + 6 y + y 2 )
x 2 + y 2 − 2 x + 4 y + 5 = 4 ( x 2 + y 2 − 8 x + 6 y + 25 ) x 2 + y 2 − 2 x + 4 y + 5 = 4 x 2 + 4 y 2 − 32 x + 24 y + 100 x 2 − 4 x 2 − 2 x + 32 x + y 2 − 4 y 2 + 4 y − 24 y + 5 − 100 = 0 − 3 x 2 + 30 x − 3 y 2 − 20 y − 95 = 0 / : ( − 3 ) x 2 − 10 x + y 2 + 3 20 ​ y + 3 95 ​ = 0 x 2 − 2 x ⋅ 5 + 5 2 − 5 2 + y 2 + 2 y ⋅ 3 10 ​ + ( 3 10 ​ ) 2 − ( 3 10 ​ ) 2 = − 3 95 ​
( x − 5 ) 2 + ( y + 3 10 ​ ) 2 = − 3 95 ​ + 9 100 ​ + 25 ( x − 5 ) 2 + ( y + 3 10 ​ ) 2 = − 9 285 ​ + 9 100 ​ + 9 225 ​ ( x − 5 ) 2 + ( y + 3 10 ​ ) 2 = 9 40 ​ i t ′ s t h e c i rc l e ce n t er : ( 5 ; − 3 10 ​ ) r a d i u s : r = 9 40 ​ ​ = 9 ​ 4 ⋅ 10 ​ ​ = 3 2 10 ​ ​

Answered by Anonymous | 2024-06-10

The curve traced by point P is a circle with the equation ( x − 5 ) 2 + ( y + 3 10 ​ ) 2 = 9 40 ​ . This circle is centered at (5, -10/3) and has a radius of 3 2 10 ​ ​ .
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Answered by Anonymous | 2024-12-23