Question

find the equation of locus of a point which is the distance of 8 units from 4, - 3​

Answer 1

it goes positive to negative so it's distance was at 8 units 0 is also the unit wants

Answer 2

The equation of the locus is

Given :

A point which is the distance of 8 units from (4, - 3)

To find :

The equation of the locus of the point

Solution :

Solution :Step 1 of 2 :

Write down the given point

Here the given point is (4, - 3)

Step 2 of 2 :

Find the locus of the point

Let (h, k) be the given point

By the given condition

$\displaystyle \sf{ \sqrt{ {(h - 4)}^{2} + {(k + 3)}^{2} } = 8 }$

$\displaystyle \sf{ \implies {(h - 4)}^{2} + {(k + 3)}^{2} = {8}^{2} }$

$\displaystyle \sf{ \implies {h}^{2} - 8h + 16 + {k}^{2} + 6k + 9 = 64 }$

$\displaystyle \sf{ \implies {h}^{2} + {k}^{2} - 8h + 6k - 39 = 0 }$

Hence the required locus of the point is

$\displaystyle \sf{ {x}^{2} + {y}^{2} - 8x + 6y - 39 = 0 }$