Does the point (-2.5, 3.5) lie inside,outside | vedclass

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(A) The equation of the given circle is $x^{2}+y^{2}=25$. This is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$,where the centre $(h, k) = (0, 0)$ and the r

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Inside the circle

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(A) The equation of the given circle is $x^{2}+y^{2}=25$. This is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$,where the centre $(h, k) = (0, 0)$ and the radius $r = \sqrt{25} = 5$. To determine the position of the point $P(-2.5, 3.5)$ relative to the circle,we calculate the distance $d$ from the centre $(0, 0)$ to the point $P$: $d = \sqrt{(-2.5 - 0)^{2} + (3.5 - 0)^{2}}$ $d = \sqrt{6.25 + 12.25}$ $d = \sqrt{18.5} \approx 4.3$. Since $d \approx 4.3 Therefore,the point $(-2.5, 3.5)$ lies inside the circle.

Does the point $(-2.5, 3.5)$ lie inside,outside,or on the circle $x^{2}+y^{2}=25$?

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