The equation of given circle is $x^2 + y^2$ = 25 ⇒ $(x-0)^2 + (y-0)^2$ = $(5)^2$ Comparing it with $(x-h)^2 + (y-k)^2$ = $r^2$, we get h = 0, k = 0, and r = 5 Now, distance of the point (–2.5, 3.5) from the centre (0, 0) = $\sqrt{(0+2.5)^2+(0-3.5)^2}$ = $\sqrt{6.25+12.25}$ = $\sqrt{18.5}$ = 4.3 < 5 $Thus the point (–2.5, 3.5) lies inside the circle.