Question

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?

Answer 1

`C (1,-3/2)` and `r=5/2`

Explanation:

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Let's create a figure we can use to solve this problem:

Point `P (x.y)` is connected to the origin `O (0,0)` and point `A (2,-3)`.

`(OP)^2+(AP)^2=19`

We can use the distance formula to calculate the lengths of `OP`and `AP`. The formula is:

`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` or

`d^2=(x_2-x_1)^2+(y_2-y_1)^2`

`(OP)^2=(x-0)^2+(y-0)^2`

`(AP)^2=(x-2)^2+(y+3)^2`

`(OP)^2=x^2+y^2` `color(red)(Equation-1)`

`(AP)^2=x^2-4x+4+y^2+6y+9` `color(red)(Equation-2)`

Now, we add `color(red)(Equation-1)` and `color(red)(Equation-2)` together:

`(OP)^2+(AP)^2=2x^2+2y^2-4x+6y+13=19`

`2x^2+2y^2-4x+6y=6`

Let's divide the entire equation by `2`:

`x^2+y^2-2x+3y=3`

We can rewrite this equation by adding and subtracting constants and grouping the `x` terms together and `y` terms together. This will allow us to complete two squares:

`(x^2-2x+1)+(y^2+3y+9/4)-1-9/4=3`

`(x-1)^2+(y+3/2)^2=3+1+9/4`

`(x-1)^2+(y+3/2)^2=25/4`

This is the equation of a circle in the form of:

`(x-h)^2+(y-k)^2=r^2`

where `h` and `k` are the `x` and `y` coordinates of its center and `r` is its radius.

Our equation is that of a circle, which is the locus of point `P (x,y)`, with Center `C (1,-3/2)` and radius `r=5/2`