Question

Find the locus of the point which
is at a constant distance of 5 units
from (4,-3)​

Answer 1

Answer:

$x^{2} +y^{2} -8x+6y=0$

Step-by-step explanation:

Let A (4,-3) and P(x, y) be a point on the locus

Given Condition is, PA = 5

⇒$PA^{2}$ = 25          ⇒$(x-4)^{2} +(y+3)^{2}=25$

⇒$x^{2} +y^{2} -8x+6y=0$

∴Locus of P is $x^{2} +y^{2} -8x+6y=0$