A moving point has its distance from (1,3) always one-third of its distance from (8,2). Find the equation of its Locus.

A moving point has its distance from (1,3) always one-third of its distance from (8,2).
Find the equation of its Locus.

My equation displays a circle formed by the loci, I don't know if it's right. Please help me find the equation, Thank you in advance to those that can answer.

• Then what is the equation of the circle you have found? – peterwhy Jan 23 '14 at 14:39
• I got two different equations of its locus, both of which does not meet the condition on the above problem. the first one is ====>8/9x^2 -149/9x +8/9y^2 - 14/3y+602/9=0 the second one is ===> -1/3x^2 +38/3x + 106/3 - 1/3y^2 +10/3y=0 – Tara Salvador Jan 23 '14 at 14:43
• We get $\sqrt{(x-1)^2+(y-3)^2}=\frac{1}{3}\sqrt{(x-8)^2+(y-2)^2}$. Square both sides, multiply through by $9$, simplify. – André Nicolas Jan 23 '14 at 14:56

\begin{align*}3^2\left[(x-1)^2+(y-3)^2\right] =& (x-8)^2 + (y-2)^2\\ 9\left(x^2-2x+y^2-6y+10\right) =& x^2-16x+y^2-4x+68\\ 8x^2-2x+8y^2-50y+22 =& 0\\ x^2-\frac14x+y^2-\frac{25}4y+\frac{11}4=&0\\ \left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\frac{225}{32}\\ \left(x-\frac{1}{8}\right)^2+\left(y-\frac{25}8\right)^2=&\left(\frac{15\sqrt2}{8}\right)^2 \end{align*}
• The fourth line to the fifth line is completing square, which is to convert $$x^2+bx = x^2+2\cdot\frac b2x+\left(\frac{b}2\right)^2 - \frac{b^2}{4}=\left(x+\frac{b}{2}\right)^2 - \frac{b^2}{4}.$$ From the fourth line, complete square once with variable $x$ and once with $y$, then move all constants to the right hand side. This is to make the quadratic equation to the form $$(x-a)^2+(y-b)^2 = r^2.$$ – peterwhy Jan 23 '14 at 15:40
• You can read the coordinates of centre and the radius from the equation of circle, if you convert the equation to $(x-a)^2+(y-b)^2=r^2$ form. Use a pair of compasses to plot. – peterwhy Jan 24 '14 at 11:51