Position of a point relative to a circle

Given the equation of a circle, and the coordinates of a point, how will we come to know that the point lies inside, outside or on the circle?

circle, position of a point

This lesson will answer this very question.

To determine the position of a given point with respect to a circle, all we need to do is to find the distance between the point and the center of the circle, and compare it with the circle’s radius.

If the distance is greater than the radius, the point lies outside. If it’s equal to the radius, the point lies on the circle. And if it’s less than the radius, you guessed it right, the point will lie inside the circle. As simple as that!

Have a look at the following figure to understand this better.

circle, position of a point

To establish some formulas, let’s take the standard equation first i.e. x2 + y2 – r2 = 0

Let P(x1, y1) be the point whose position is to be determined. Then it’s distance from the center of the circle C(0, 0) will be given by CP = \( \sqrt{x^2+y^2} \)

Now, the point will lie outside, on or inside the circle when CP > r, CP = r or CP < r respectively. By rearrangement of terms, we get the required conditions as

Outside: x12 + y12 – r2 > 0

On: x12 + y12 – r2 = 0

Inside: x12 + y12 – r2 < 0

That is, substitute the coordinates of the point in the circle’s equation. If the expression is positive, then the point lies outside circle, if zero, then on the circle, and if negative then inside the circle. Neat.

Now let’s move on to the general equation: x2 + y2 + 2gx + 2fy + c = 0. Not much of a difference here, except that the center is (-g, -f) and the radius \( \sqrt{g^2 +f^2 – c} \) . In this case the distance of the point from the center will be \( \sqrt{(x+g)^2+(y+f)^2}\)

Comparing this with the radius, we get the conditions as:

Outside: x12 + y12 + 2gx1 + 2fy1 + c > 0

On: x12 + y12 + 2gx1 + 2fy1 + c = 0

Inside: x12 + y12 + 2gx1 + 2fy1 + c < 0

Hmm. We get something similar again. The expression whose sign we’re checking is the one on the left hand side of the circle’s equation, in which the coordinates (x1, y1) are substituted in place of (x, y).

In fact, that expression will be of much interest to us, throughout our journey in conic sections. So we’ll give it a name: S1

Let S = x2 + y2 + 2gx + 2fy + c, and P be the point (x1, y1). Then S1 = x12 + y12 + 2gx1 + 2fy1 + c

And the position of the point P is determined by checking the sign of S1, as seen above!

To summarize: The point lies outside, inside or on the circle when S1 > 0, S1 < 0 or S1 = 0 respectively.

That’s it for the current lesson, see you in the next with some examples.

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