Simple 7

Geometry Level 3

Find the position of the point ( 1 , 2 ) (1,2) with respect to the circle x 2 + y 2 − 2 x − 3 y − 4 = 0 x^2+y^2-2x-3y-4 = 0 .

Inside On the circle Outside

Yash Dev Lamba by Yash Dev Lamba , 23, India

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2 solutions

Krishna Ramesh
Feb 2, 2016

just directly substitute the point in the equation of the circle.

the value is less than 0, hence the point lies inside the circle

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Roger Erisman
Jan 22, 2016

Completing the square:

x^2 - 2x + 1 + y^2 - 3y + 2.25 = 4 + 1 + 2.25

(x - 1)^2 + (y - 1.5)^2 = 7.25

center at (1,1.5)

(1,2) is 0.5 away from center, but circle itself is sqrt(7.25) = 2.6926 away so point is inside circle

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