Question

A circle is centered at C(−1,−3) and has a radius of 6.Where does the point P(-6,-6) lie?

Answer 1

The equation of circle is

$(x-h)^{2}  + (y-k)^{2} = r^{2}$.   -------- (h, k) is the center of the circle.

The point is inside the circle if the distance from it to the centre is less than the radius. Symbolically, this is

$\sqrt{({x - h})^{2} +{(y - k)}^{2}} < r$

and point is outside the circle if the distance from it to the centre is great than the radius. Symbolically, this is

$\sqrt{({x - h})^{2} +{(y - k)}^{2}} > r$

and if this distance is equal to r then it lies on the cirlce.Symbolically, this is

$\sqrt{({x - h})^{2} +{(y - k)}^{2}} = r$

now by putting the value of (x , y) and (h,k)

$\sqrt{(-6 - -1)^{2} + (-6 - -3)^{2}}$

$=\sqrt{25 + 9}$

$=\sqrt{34}$

since our r is 6 and

=5.8 < 6

So the point (-6 , -6) is inside the circle.

Answer 2

Answer:

inside the circle

Explanation:

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