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A point moves so its distance from $(3,2)$ is half its distance from the line $2x+3y=1$. The locus is:
Aa circle of radius 3 centred at $(3,2)$
Bnot a circle, because an $xy$ term survives
Ca circle, but with neither centre nor radius
Da straight line running parallel to $2x+3y=1$
Answer & Solution
Correct answer: B. not a circle, because an $xy$ term survives
1. Set up the condition: $\sqrt{(x-3)^2+(y-2)^2}=\dfrac12\cdot\dfrac{|2x+3y-1|}{\sqrt{2^2+3^2}}$.
2. Squaring both sides and clearing the denominator $13$ gives a second-degree equation.
3. Simplifying leads to $48x^2+43y^2-308x-202y-12xy+675=0$.
4. A circle needs equal coefficients on $x^2$ and $y^2$ and no $xy$ term. Here $48\neq43$ and $-12xy$ is present.
5. So the locus is not a circle, and it has neither a centre nor a radius.
_Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 8: Coordinate Geometry II_