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:

\((x-(-2))^2+(y-3)^2=(x-4)^2+(y-(-1))^2\).

Expand and simplify the resulting equation:

\(\begin{aligned} (x+2)^2+(y-3)^2&=(x-4)^2+(y+1)^2 \\ x^2+2x+4+y^2-6y+9&=x^2-8x+16+y^2+2y+1 \\ 10x-8y-4&=0. \end{aligned}\)

The equation of the locus of a moving point \(P\) is:

\(10x-8y-4=0\).

\(A(2,0)\) 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:

\(\begin{aligned} \dfrac{AP}{PB}&=\dfrac{1}{2} \\ \dfrac{(x-2)^2+(y-0)^2}{(x-0)^2+(y+2)^2}&=\dfrac{1}{2^2}. \end{aligned}\)

Expand and simplify the resulting equation:

\(\begin{aligned} 4[(x-2)^2+y^2]&=x^2+(y+2)^2 \\ 4(x^2-4x+4)+4y^2&=x^2+(y^2+4y+4) \\ 4x^2-16x+16+4y^2&=x^2+y^2+4y+4 \\ 3x^2+3y^2-16x-4y+12&=0. \end{aligned}\)

The equation of the locus of a moving point \(P\) is:

\(3x^2+3y^2-16x-4y+12=0\).

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:

\((x-(-2))^2+(y-3)^2=(x-4)^2+(y-(-1))^2\).

Expand and simplify the resulting equation:

\(\begin{aligned} (x+2)^2+(y-3)^2&=(x-4)^2+(y+1)^2 \\ x^2+2x+4+y^2-6y+9&=x^2-8x+16+y^2+2y+1 \\ 10x-8y-4&=0. \end{aligned}\)

The equation of the locus of a moving point \(P\) is:

\(10x-8y-4=0\).

\(A(2,0)\) 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:

\(\begin{aligned} \dfrac{AP}{PB}&=\dfrac{1}{2} \\ \dfrac{(x-2)^2+(y-0)^2}{(x-0)^2+(y+2)^2}&=\dfrac{1}{2^2}. \end{aligned}\)

Expand and simplify the resulting equation:

\(\begin{aligned} 4[(x-2)^2+y^2]&=x^2+(y+2)^2 \\ 4(x^2-4x+4)+4y^2&=x^2+(y^2+4y+4) \\ 4x^2-16x+16+4y^2&=x^2+y^2+4y+4 \\ 3x^2+3y^2-16x-4y+12&=0. \end{aligned}\)

The equation of the locus of a moving point \(P\) is:

\(3x^2+3y^2-16x-4y+12=0\).