Question

The locus of point which is at a distance 5 unit from (–2, 3) is​

Answer 1

Let the point be P(h,k)

So distance between P(h,k) and Q(-2,3) = 5 units

We know that , distance between two points is given by :-

$\underline{ \boxed{ \bf \large \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} } = d }}$

$\sf where (x_1, y_1) \: and \: (x_2 , y_2) \: are \: two \: points \: and \: \bf \: d \sf \: is \: distance \: between \: them$

So distance between point P(h,k) and Q(-2,3) :-

$\sf \implies\sqrt{ {( - 2 - h)}^{2} + {(3-k)}^{2} } = 5$

$\sf \implies\sqrt{ { { (- 1)}^{2} ( 2 + h)}^{2} + {(3-k)}^{2} } = 5$

$\sf \implies[ {( 2 + h)^{2} + {(3-k)}^{2} } ]^{ \frac{1}{2} } = {5}$

$\sf \implies( 2 + h)^{2} + {(3-k)}^{2} = {5}^{2}$

$\sf \implies4 + {h}^{2} + 4h + 9 + {k}^{2} - 6k= 25$

$\sf \implies13 + {h}^{2} + 4h + {k}^{2} - 6k = 25$

$\sf \implies {h}^{2} + 4h + {k}^{2} - 6k - 25 + 13 = 0$

$\sf \implies {h}^{2} + 4h + {k}^{2} - 6k - 12 = 0$

Now replace h from x and k from y to get locus.

$\sf \implies {x}^{2} + 4x + {y}^{2} - 6y - 12 = 0$

Locus of point which is at a distance 5 unit from (–2, 3) is circle of radius 5 and centre (-2,3).