Does the point #(4,-3)# lie on the circle #(x+4)^2+(y-3)^2=49#?

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Answer 1 · 2017-11-27T12:18:28

No, it does't lie on the circle.

The equation of a circle is given by:
#(x-a)^2+(y-b)^2=r^2#, where:

Using the given information we can get:
#(x+4)^2+(y-3)^2=7^2=49#

Using the the point #(4,-3)=>(x,y)# we get:
#(4+4)^2+(-3-3)^2=8^2+(-6)^2=64+36=100#

#sqrt(100)=10!=7#

Answer 2 · 2017-11-27T12:21:30

No, it does not.

For the point #(4,-3)# to lie on the circle, its distance from the centre must be exactly #7# units.

To find the distance between the point and the centre, use Pythagoras Theorem,

Distance#=sqrt((4-(-4))^2+(-3-3)^2)#
#color(white)(xxxx/x.)=sqrt100#
#color(white)(xxxx/x.)=10#

Since the distance from of the point from the centre is more than #7#, the point lies outside of the circle and does not lie on the circle.

Check:

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