Locus of moving point \(P(x,y)\) which distance is constant from a single fixed point \(A(x_1,y_1)\) is:
\((x-x_1)^2+(y-y_1)^2=r^2\)
Locus of moving point \(P(x,y)\) which distance from two fixed points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is always constant in the ratio \(m:n\) is:
\(\dfrac{(x-x_1)^2+(y-y_1)^2}{(x-x_2)^2+(y-y_2)^2}=\dfrac{m^2}{n^2}\)
Locus of moving point \(P(x,y)\) which distance from two fixed points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is the same is:
\((x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2\)
Find the equation of the locus of a moving point \(P(x,y)\) so that its distance is always \(5\) units from a fixed point \(Q(2,4)\).
The formula involved is (a fixed point):
\((x-x_1)^2+(y-y_1)^2=r^2\).
Substitute the value of the coordinate \(Q(2,4)\) and the distance of \(5\) units into the formula:
\((x-2)^2+(y-4)^2=5^2\).
Expand and simplify the resulting equation:
\(\begin{aligned} x^2-4x+4+y^2-8y+16&=25 \\ x^2+y^2-4x-8y-5&=0. \end{aligned}\)
The equation of the locus of a moving point \(P\) is:
\(x^2+y^2-4x-8y-5=0\).
Find the equation of the locus of the moving point \(P(x,y)\) so that its distance from point \(A(-2,3)\) and point \(B(4,-1)\) is the same.
The formula involved is (two fixed point):
\((x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2\).
Substitute the value of the coordinates \(A(-2,3)\) and \(B(4,-1)\) into the formula:
\((x-(-2))^2+(y-3)^2=(x-4)^2+(y-(-1))^2\).
\(\begin{aligned} (x+2)^2+(y-3)^2&=(x-4)^2+(y+1)^2 \\ x^2+2x+4+y^2-6y+9&=x^2-8x+16+y^2+2y+1 \\ 10x-8y-4&=0. \end{aligned}\)
\(10x-8y-4=0\).
\(A(2,0)\) and \(B(0,-2)\) are two fixed points. Point \(P\) moves in a ratio such that \(AP:PB=1:2\). Find the equation of the locus of the moving point \(P\).
The formula involved is (two fixed point, ratio):
\(\dfrac{(x-x_1)^2+(y-y_1)^2}{(x-x_2)^2+(y-y_2)^2}=\dfrac{m^2}{n^2}\).
Substitute the value of the coordinates \(A(2,0)\) and \(B(4,-1)\) with ratio \(AP:PB=1:2\) into the formula:
\(\begin{aligned} \dfrac{AP}{PB}&=\dfrac{1}{2} \\ \dfrac{(x-2)^2+(y-0)^2}{(x-0)^2+(y+2)^2}&=\dfrac{1}{2^2}. \end{aligned}\)
\(\begin{aligned} 4[(x-2)^2+y^2]&=x^2+(y+2)^2 \\ 4(x^2-4x+4)+4y^2&=x^2+(y^2+4y+4) \\ 4x^2-16x+16+4y^2&=x^2+y^2+4y+4 \\ 3x^2+3y^2-16x-4y+12&=0. \end{aligned}\)
\(3x^2+3y^2-16x-4y+12=0\).
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