Interaction of a Point with a Circle

When a point is near a circle, many interesting questions come up about how the point and the circle are related. First, we often want to know whether the point lies inside, outside, or on the circle. This depends on how far the point is from the center of the circle compared to the radius.

Next, we might ask: what is the shortest distance from the point to the circle? What is the longest distance from the point to the circle? These distances help us understand how close or far the point is from different parts of the circle.

There are also deeper ideas. One of them is the power of a point with respect to a circle. This gives us a number that tells us how the point is related to the circle in terms of distance. Another idea is the inverse point with respect to a circle, which comes from a special transformation in geometry called inversion.

In the sections ahead, we will carefully define and explain each of these ideas, using diagrams, formulas, and reasoning that make their meanings clear.

Relative Location of a Point with Respect to a Circle

Suppose we are given a circle and a point in the plane. A natural question is whether the point lies inside the circle, on the circle, or outside it. While one could draw a diagram and check visually, this is not reliable or efficient. A better approach is to use algebra.

Let the equation of the circle be

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \tag{1} \]

and let the point be \( P(x_1, y_1) \). The center of the circle is

\[ C(-g, -f) \quad \text{and the radius is} \quad r = \sqrt{g^2 + f^2 - c}. \]

We now measure the distance \( CP \), the distance from the point \( P \) to the center \( C \). This is given by

\[ CP = \sqrt{(x_1 + g)^2 + (y_1 + f)^2}. \]

To determine the position of the point relative to the circle, we compare this distance \( CP \) with the radius \( r \):

• If \( CP < r \), then the point lies inside the circle.
• If \( CP = r \), then the point lies on the circle.
• If \( CP > r \), then the point lies outside the circle.

To avoid square roots, we square both sides. For example, the point is inside the circle if and only if

\[ (x_1 + g)^2 + (y_1 + f)^2 < g^2 + f^2 - c. \]

Expanding the left side:

\[ x_1^2 + 2gx_1 + g^2 + y_1^2 + 2f y_1 + f^2 < g^2 + f^2 - c \]

Cancelling \( g^2 + f^2 \) from both sides, we get

\[ x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c < 0. \]

This expression is obtained by substituting \( (x_1, y_1) \) into the left-hand side of the circle's equation. We write this compactly as

\[ S_1 < 0, \]

where \( S_1 \) denotes the result of substituting \( (x_1, y_1) \) into the circle equation (1).

Similarly, we can prove that when a point is outside then \(S_1>0\) taking \(CP > r\). For the point to be on the circle, the jpoint simply satisfies the circle.

Thus, we summarize:

\[ \begin{aligned} \text{If } S_1 < 0 &\implies \text{Point lies inside the circle}, \\ \text{If } S_1 = 0 &\implies \text{Point lies on the circle}, \\ \text{If } S_1 > 0 &\implies \text{Point lies outside the circle}. \end{aligned} \]

This provides a direct algebraic test for the position of any point with respect to a given circle.

Every point that lies inside the circle, when substituted into the left-hand side of the circle's equation

\[ x^2 + y^2 + 2gx + 2fy + c, \]

gives a negative value. Conversely, every point outside the circle gives a positive value when substituted into the same expression. Points on the circle make the expression exactly zero.

This observation shows that the circle acts as a boundary between two distinct regions of the plane:

• The set of all points for which

  \[ x^2 + y^2 + 2gx + 2fy + c < 0 \]

  lies inside the circle.

• The set of all points for which

  \[ x^2 + y^2 + 2gx + 2fy + c > 0 \]

  lies outside the circle.

• The set of all points for which

  \[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

  lies exactly on the circle.

Thus, the circle naturally divides the plane into two regions: an interior and an exterior, separated by the curve itself.

Power of a Point with Respect to a Circle

Let a circle be given by the equation

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \tag{1} \]

and let \( P(x_1, y_1) \) be any point in the plane. The power of the point \( P \) with respect to this circle is defined as the value obtained by substituting the coordinates of \( P \) into the left-hand side of equation (1). That is, the power of \( P \) is

\[ \Pi(P) = x_1^2 + y_1^2 + 2g x_1 + 2f y_1 + c. \tag{2} \]

This expression is denoted as \( S_1 \), and we have already seen that:

\[ \begin{aligned} S_1 < 0 &\implies P \text{ lies inside the circle}, \\ S_1 = 0 &\implies P \text{ lies on the circle}, \\ S_1 > 0 &\implies P \text{ lies outside the circle}. \end{aligned} \]

However, the significance of this value goes beyond merely determining the location. It represents a quantity with geometric meaning.

Let the center of the circle be \( C(-g, -f) \) and the radius be

\[ r = \sqrt{g^2 + f^2 - c}. \tag{3} \]

The distance from the point \( P \) to the center \( C \) is

\[ CP = \sqrt{(x_1 + g)^2 + (y_1 + f)^2}. \tag{4} \]

Squaring both sides and comparing with \( r^2 \), we define the power as:

\[ \Pi(P) = CP^2 - r^2. \tag{5} \]

Substituting (3) and (4) into (5), we get:

\[ \begin{aligned} \Pi(P) &= (x_1 + g)^2 + (y_1 + f)^2 - (g^2 + f^2 - c) \\ &= x_1^2 + 2g x_1 + g^2 + y_1^2 + 2f y_1 + f^2 - g^2 - f^2 + c \\ &= x_1^2 + y_1^2 + 2g x_1 + 2f y_1 + c. \end{aligned} \]

This confirms that both definitions (geometric and algebraic) agree:

\[ \Pi(P) = S_1. \]

Hence, the power of a point is equal to the difference between the square of its distance from the center and the square of the radius:

\[ \boxed{\Pi(P) = CP^2 - r^2}. \]

Power and Secant-Secant Theorem

Let a point \( P \) lie in the plane of a circle, and let two secants be drawn from it. Suppose one secant intersects the circle at points \( S_1 \) and \( S_2 \), and the other at \( S_3 \) and \( S_4 \). Then by the Secant–Secant Theorem, we have

\[ PS_1 \cdot PS_2 = PS_3 \cdot PS_4. \]

From geometry, we know that when the point \( P \) lies outside the circle, the product of the lengths of the segments from \( P \) to the two points of intersection on any secant equals the square of the distance from \( P \) to the center minus the square of the radius. That is,

\[ PS_1 \cdot PS_2 = PS_3 \cdot PS_4 = CP^2 - r^2. \]

But the right-hand side is precisely the definition of the power of the point \( P \) with respect to the circle. Hence, when \( P \) is outside the circle,

\[ \boxed{PS_1 \cdot PS_2 = PS_3 \cdot PS_4 = \Pi(P)}. \]

Now suppose the point \( P \) lies inside the circle. Then the secants become chords intersecting inside the circle at \( P \), and the same product still holds by geometry:

\[ PS_1 \cdot PS_2 = PS_3 \cdot PS_4. \]

However, in this case,

\[ PS_1 \cdot PS_2 = PS_3 \cdot PS_4 = r^2 - CP^2 = -\Pi(P). \]

Thus, when the point lies inside the circle, the common product of the chord segments equals the negative of the power of the point. Both cases combine into the identity:

\[ PS_1 \cdot PS_2 = PS_3 \cdot PS_4 = \left| \Pi(P) \right|, \]

with the appropriate sign given by the position of \( P \) relative to the circle.

Inversion with Respect to a Circle

Let a fixed circle be given with center \( C \) and radius \( r \). For any point \( P \ne C \), its inverse point \( Q \) with respect to this circle is defined by the condition

\[ CP \cdot CQ = r^2. \]

Let \( CP = d \), so that the distance from \( C \) to \( P \) is \( d \). Then, by the definition of inversion,

\[ CQ = \frac{r^2}{d}. \]

Hence, the inverse point \( Q \) lies on the same ray as \( CP \), and its position is determined by the reciprocal relationship between \( d \) and \( CQ \).

Suppose \( P \) lies inside the circle. Then \( d < r \), and so

\[ CQ = \frac{r^2}{d} = r \cdot \left( \frac{r}{d} \right). \]

Since \( \frac{r}{d} > 1 \), it follows that \( CQ > r \), meaning that \( Q \) lies outside the circle.

Similarly, if \( P \) lies outside the circle, then \( d > r \), so \( \frac{r}{d} < 1 \), hence \( CQ < r \), and therefore \( Q \) lies inside the circle.

If \( P \) lies on the circle, then \( d = r \), and we have \( CQ = \frac{r^2}{r} = r \), so \( CQ = CP \), and the inverse point \( Q \) coincides with \( P \) itself. In this case, the point is invariant under inversion.

Now assume \( P \) lies inside the circle, so \( d < r \). Then

\[ PQ = CQ - CP = \frac{r^2}{d} - d = \frac{r^2 - d^2}{d}. \]

This means that \( P \) divides the segment \( CQ \) internally in the ratio

\[ CP : PQ = d : \frac{r^2 - d^2}{d} = d^2 : (r^2 - d^2). \]

On the other hand, if \( P \) lies outside the circle, then \( d > r \), and now

\[ PQ = CP - CQ = d - \frac{r^2}{d} = \frac{d^2 - r^2}{d}. \]

So \( P \) again divides the segment \( CQ \) but externally, but now the ratio is

\[ CP : PQ = d : \frac{d^2 - r^2}{d} = d^2 : (d^2 - r^2), \]

Thus, in all cases, the point \( P \) divides the segment \( CQ \) in the ratio

\[ \boxed{d^2 : (r^2 - d^2)}, \]

internally if \( P \) lies inside the circle and externally if \( P \) lies outside (as then \(d>r\) the ratio becomes negative implying external division).

Let the circle be given by

\[ x^2 + y^2 = r^2, \]

whose center is the origin \( C = (0, 0) \). Let \( P(x, y) \) be any point in the plane, distinct from \( C \), and let \( d = CP = \sqrt{x^2 + y^2} \) be the distance from the center to the point \( P \). Let \( Q \) be the inverse point of \( P \) with respect to the given circle. By the definition of inversion,

\[ CP \cdot CQ = r^2 \implies CQ = \frac{r^2}{d}. \]

From earlier reasoning, the point \( P \) divides the segment \( CQ \) in the ratio

\[ CP : PQ = d^2 : (r^2 - d^2). \]

Then, by the division formula, we write

\[ P = \frac{(r^2 - d^2) \cdot O + d^2 \cdot Q}{r^2}, \]

where \( O = (0, 0) \) is the origin. Simplifying this gives

\[ P = \frac{d^2}{r^2} \cdot Q \implies Q = \frac{r^2}{d^2} \cdot P. \]

But since \( d^2 = x^2 + y^2 \), we substitute to obtain

\[ Q = \left( \frac{r^2 x}{x^2 + y^2}, \, \frac{r^2 y}{x^2 + y^2} \right). \]

This is the explicit coordinate formula for the inverse point of \( P(x, y) \) with respect to the circle \( x^2 + y^2 = r^2 \).

Inversion of a Line in a Circle

Let the circle of inversion be

\[ x^2 + y^2 = r^2, \]

with center at the origin \( O = (0, 0) \). Consider a vertical line

\[ x = d, \]

and let a point on this line be \( P(d, \beta) \), where \( \beta \in \mathbb{R} \). Then \( P \) lies on the given line, and its distance from the origin is

\[ OP = \sqrt{d^2 + \beta^2}. \]

The inverse point \( Q \) of \( P \) with respect to the circle is given by the inversion formula:

\[ Q = \left( \frac{r^2 d}{d^2 + \beta^2}, \, \frac{r^2 \beta}{d^2 + \beta^2} \right). \]

Let

\[ x = \frac{r^2 d}{d^2 + \beta^2}, \quad y = \frac{r^2 \beta}{d^2 + \beta^2}. \tag{1, 2} \]

We now square both expressions and add to eliminate \( \beta \):

\[ x^2 + y^2 = \frac{r^4 d^2}{(d^2 + \beta^2)^2} + \frac{r^4 \beta^2}{(d^2 + \beta^2)^2} = \frac{r^4 (d^2 + \beta^2)}{(d^2 + \beta^2)^2} = \frac{r^4}{d^2 + \beta^2}. \tag{3} \]

From this, it follows that

\[ d^2 + \beta^2 = \frac{r^4}{x^2 + y^2}. \tag{4} \]

Substitute this into equation (1):

\[ x = \frac{r^2 d}{d^2 + \beta^2} = r^2 d \cdot \frac{x^2 + y^2}{r^4} = \frac{d}{r^2} (x^2 + y^2). \tag{5} \]

Multiplying both sides by \( r^2 \), we obtain the equation:

\[ r^2 x = d(x^2 + y^2) \implies x^2 + y^2 - \frac{r^2}{d} x = 0. \tag{6} \]

This is the equation of a circle. To write it in standard form, complete the square in \( x \):

\[ x^2 - \frac{r^2}{d} x + y^2 = 0 \implies \left(x - \frac{r^2}{2d} \right)^2 + y^2 = \left( \frac{r^2}{2d} \right)^2. \]

Hence, the locus of inverse points is a circle with center

\[ \left( \frac{r^2}{2d}, \, 0 \right) \]

and radius

\[ \frac{r^2}{2d}. \]

This circle passes through the origin, since setting \( x = 0, y = 0 \) satisfies equation (6).

Therefore, we have shown that when each point on the line \( x = d \) is inverted with respect to the circle \( x^2 + y^2 = r^2 \), the set of all inverse points traces out a circle that passes through the center of inversion. Hence:

The inversion of a straight line not passing through the center is a circle passing through the center.

A similar analysis shows the converse: the inversion of a circle passing through the center is a straight line not passing through the center, completing the duality between lines and circles under inversion.

Minimum and Maximum Distance of a Point from a Circle

Let a circle be given with center \( O \) and radius \( r \). Let \( P(x_1, y_1) \) be any fixed point in the plane, and let \( Q \) be any point on the circle. The distance \( PQ \) varies as \( Q \) moves along the circle. We are interested in finding the minimum and maximum values of \( PQ \), that is, the shortest and longest distances from the point \( P \) to any point \( Q \) on the circle.

Let \( d = OP \) be the distance from the point \( P \) to the center of the circle. We consider two cases based on the location of \( P \) relative to the circle.

Assume first that \( P \) lies outside the circle, so that \( d > r \). Consider the straight line segment \( \overline{OP} \) joining the center \( O \) to the point \( P \). This line intersects the circle at exactly two points, say \( Q_1 \) and \( Q_2 \). One of these points, \( Q_1 \), lies on the near side of the circle, toward \( P \); the other, \( Q_2 \), lies on the far side. Since \( Q_1 \) lies between \( O \) and \( P \), the distance \( PQ_1 = OP - r = d - r \), which is the shortest possible distance from \( P \) to the circle. Likewise, \( Q_2 \) lies beyond the center in the direction of the line \( \overline{OP} \), so \( PQ_2 = OP + r = d + r \), which is the maximum.

Hence, when \( P \) is outside the circle:

\[ \text{Minimum distance from } P \text{ to the circle} = d - r, \\ \text{Maximum distance from } P \text{ to the circle} = d + r. \]

Now consider the case when \( P \) lies inside the circle, so \( d < r \). The same line \( \overline{OP} \) intersects the circle at two points \( Q_1 \) and \( Q_2 \) again, this time lying outside the point \( P \) in both directions. Still, the shortest distance from \( P \) to any point \( Q \) on the circle is attained when \( Q = Q_1 \), the point on the circle closest to \( P \) along the line \( \overline{OP} \). This time:

\[ PQ_1 = r - d, \quad PQ_2 = r + d. \]

Thus, regardless of whether \( P \) is inside or outside the circle, the minimum and maximum distances from the point \( P \) to the circle are given by:

\[ \boxed{\min PQ = |d - r|, \quad \max PQ = d + r}, \]

where \( d = OP \). This expression holds uniformly, since it adjusts naturally to both \( d > r \) and \( d < r \).