Question

The position of the point P(2, 6) w.r.t to the circle

x^2
+ ^2
− 2 − 3 − 1 = 0 is​

Answer 1

Answer:

ANSWER

We have x

2

+y

2

−4x+2y−11=0 or S = 0 where S=x

2

+y

2

−4x+2y−11

For the point (1, 2) we have S

1

=1

2

+2

2

−4×1+2×2−11<0

For the point (6, 0) we have S

2

=6

2

+0

2

−4×6+2×0−11>0

Hence the point (1, 2) lies inside the circle and the point (6, 0) lies outside the circle