# A point moves in a manner that the sum of the squares of its distance from thr origin and the point (2,-3) is always 19.show that the locus of moving point is a circle.find the equation to the circle?

Jul 31, 2018

${x}^{2} + {y}^{2} - 2 x + 3 y - 3 = 0$.

#### Explanation:

Let the variable point be $P \left(x , y\right)$ and the given fixed points be

$O \left(0 , 0\right) \mathmr{and} A \left(2 , - 3\right)$.

It is given that, $P {O}^{2} + P {A}^{2} = 19$.

Using the distance formula, we have,

$\therefore \left\{{\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2}\right\} + \left\{{\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2}\right\} = 19$.

$\therefore 2 {x}^{2} + 2 {y}^{2} - 4 x + 6 y + 4 + 9 = 19 , \mathmr{and} ,$

${x}^{2} + {y}^{2} - 2 x + 3 y - 3 = 0$.

Rewriting after completing squares,

${\left(x - 1\right)}^{2} + {\left(y + \frac{3}{2}\right)}^{2} = 3 + 1 + \frac{9}{4} = \frac{25}{4} = {\left(\frac{5}{2}\right)}^{2}$.

This exhibits that the locus of the point is a circle having

centre at $\left(1 , - \frac{3}{2}\right)$ and radius $\frac{5}{2}$.

It may be interesting to note that the centre of the circle is the

mid-point of $O A$.