Pair of Tangents

Suppose there is a point \( P(x_1, y_1) \) located outside a given circle. From geometry, it is well known that exactly two tangents can be drawn from such a point to the circle. Let these tangents touch the circle at points \( Q \) and \( R \). These points of contact lie on the circle, and the lines \( PQ \) and \( PR \) are equal in length and meet at \( P \), forming an angle between them. The line joining the points of contact, namely the segment \( QR \), is called the chord of contact.

We are interested in many things here like length of these tangents, equation of these tangents, angle between these tangents, equation of chord of contact and many other things which arises out of this geommetry.

Length of a Tangent Drawn from an External Point

Let \( P(x_1, y_1) \) be a point located outside the circle

\[ S: \quad x^2 + y^2 + 2gx + 2fy + c = 0. \]

Let \( C = (-g, -f) \) be the center of the circle, and let \( PT \) be a tangent drawn from point \( P \) to the circle, touching the circle at point \( T \). Join the points \( P \) and \( C \), forming triangle \( \triangle PCT \).

Since \( PT \) is a tangent to the circle and \( CT \) is the radius, the triangle \( PCT \) is a right-angled triangle at \( T \). Thus, by the Pythagorean Theorem:

\[ PT^2 = PC^2 - CT^2. \]

The distance \( PC \) is:

\[ PC^2 = (x_1 + g)^2 + (y_1 + f)^2, \]

and the radius squared is:

\[ CT^2 = g^2 + f^2 - c. \]

So we get:

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

Simplifying:

\[ PT^2 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + g^2 + f^2 - g^2 - f^2 + c = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c. \]

Therefore,

\[ PT = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}. \]

This expression is exactly the value obtained when we substitute the point \( (x_1, y_1) \) into the equation of the circle \( S \), and evaluate its left-hand side:

\[ S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c. \]

We define this as \( S_1 \), the value of the circle's expression at the point \( (x_1, y_1) \). Hence, the length of the tangent from point \( P \) to the circle is:

\[ \boxed{PT = \sqrt{S_1}}, \]

provided \( S_1 > 0 \), which is the condition for the point to lie outside the circle.

Equation of Pair of Tangents from a Point Outside a Circle

Let \( P(x_1, y_1) \) be a point outside the circle. Let the circle be

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

which is centered at the origin and has radius \( a \). We are interested in finding the equations of the tangents from \( P \) to the circle. Although geometry tells us there should be two such tangents when the point lies outside the circle, we do not assume this here. Instead, we let the algebra guide us.

Suppose the tangent from \( P \) has slope \( m \). At this stage, we do not assume whether there are one or two such lines. The general equation of a line with slope \( m \) passing through point \( (x_1, y_1) \) is:

\[ y - y_1 = m(x - x_1) \implies y = mx + (y_1 - mx_1). \]

This is a candidate tangent to the circle. To ensure that it is indeed a tangent, it must satisfy the condition of tangency with the circle. We apply the known condition of tangency for a line \( y = mx + c \) to the circle \( x^2 + y^2 = a^2 \), which is:

\[ c^2 = a^2(1 + m^2). \]

Here, \( c = y_1 - mx_1 \), so:

\[ (y_1 - mx_1)^2 = a^2(1 + m^2). \]

Expanding both sides: [ y_1^2 - 2mx_1y_1 + m^2x_12 = a^2 + a^2m2. ]

Group terms:

\[ m^2(x_1^2 - a^2) - 2x_1y_1 m + (y_1^2 - a^2) = 0. \]

Thus, we arrive at a quadratic equation in \( m \)

\[ (x_1^2 - a^2)m^2 - 2x_1y_1 m + (y_1^2 - a^2) = 0. \tag{1} \]

This equation determines the possible slopes \( m_1 \) and \( m_2 \) of the tangents from \( P(x_1, y_1) \) to the circle. Whether one or two real tangents exist depends on the discriminant \( \Delta \) of this quadratic.

Compute the discriminant of (1):

\[ \Delta = (-2x_1y_1)^2 - 4(x_1^2 - a^2)(y_1^2 - a^2) = 4x_1^2y_1^2 - 4(x_1^2 - a^2)(y_1^2 - a^2). \]

Simplify:

\[ \Delta = 4\left[ x_1^2y_1^2 - (x_1^2y_1^2 - a^2x_1^2 - a^2y_1^2 + a^4) \right] = 4(a^2x_1^2 + a^2y_1^2 - a^4). \]

Factor out \( a^2 \):

\[ \Delta = 4a^2(x_1^2 + y_1^2 - a^2). \]

Now, since \( \Delta > 0 \) for two distinct real roots (i.e., two distinct tangents), this implies:

\[ x_1^2 + y_1^2 - a^2 > 0 \implies x_1^2 + y_1^2 > a^2, \]

which means that the point \( P \) lies outside the circle. This is exactly what geometry predicts. Hence, algebra agrees with the geometrical result.

Thus, the equation (1) yields the two slopes \( m_1 \) and \( m_2 \) of the tangents from point \( P \) to the circle. Once these slopes are known, the corresponding tangent lines can be written in the form:

\[ y - y_1 = m(x - x_1), \]

where \( m = m_1 \) or \( m_2 \).

Example

Find the equations of the tangents drawn from the point \( P(3, 2) \) to the circle

\[ x^2 + y^2 = 4. \]

Solution:

Let the slope of one such tangent be \( m \). Then the equation of the line through \( P(3, 2) \) with slope \( m \) is:

\[ y - 2 = m(x - 3) \implies y = mx + 2 - 3m. \]

This line is a tangent to the circle \( x^2 + y^2 = 4 \), so it must satisfy the condition of tangency:
If the equation is \( y = mx + c \), then tangency requires:

\[ c^2 = a^2(1 + m^2), \]

where \( a = 2 \) (since the circle has radius 2).

Here, \( c = 2 - 3m \), so:

\[ (2 - 3m)^2 = 4(1 + m^2). \]

Expand both sides:

\[ 4 - 12m + 9m^2 = 4 + 4m^2 \]

\[ \implies 9m^2 - 12m + 4 - 4 - 4m^2 = 0 \implies 5m^2 - 12m = 0 \]

\[ \implies m(5m - 12) = 0. \]

So the two possible slopes are:

\[ m = 0 \quad \text{or} \quad m = \frac{12}{5}. \]

Now, find the corresponding equations of the tangents.

For \( m = 0 \):

\[ y = 0x + 2 - 3(0) = 2 \implies \boxed{y = 2}. \]

For \( m = \frac{12}{5} \):

\[ y = \frac{12}{5}x + 2 - 3 \cdot \frac{12}{5} = \frac{12}{5}x + 2 - \frac{36}{5} = \frac{12}{5}x - \frac{26}{5}. \]

Thus, the second tangent is:

\[ \boxed{y = \frac{12}{5}x - \frac{26}{5}}. \]

These are the two tangents from \( (3,\ 2) \) to the circle \( x^2 + y^2 = 4 \).

Slope of one tangent is not defined

Find the equations of the tangents drawn from the point \( (2,\ 3) \) to the circle

\[ x^2 + y^2 = 4. \]

Solution: '

Let the slope of a tangent from point \( P(2,\ 3) \) be \( m \). Then the equation of the line through \( P \) with slope \( m \) is:

\[ y - 3 = m(x - 2) \implies y = mx + 3 - 2m. \]

This line is a tangent to the circle \( x^2 + y^2 = 4 \), so it must satisfy the condition of tangency:

\[ c^2 = a^2(1 + m^2), \]

with \( a = 2 \) and \( c = 3 - 2m \). Thus:

\[ (3 - 2m)^2 = 4(1 + m^2). \]

Expanding both sides:

\[ 9 - 12m + 4m^2 = 4 + 4m^2 \implies 9 - 12m = 4 \implies 5 = 12m \implies m = \frac{5}{12}. \]

We appear to get only one solution for \( m \), suggesting that there is only one tangent. But this cannot be true, since the point \( (2,\ 3) \) lies outside the circle \( x^2 + y^2 = 4 \), and from geometry we know there must be two distinct tangents.

The resolution is that the second tangent has undefined slope, i.e., it is vertical. A vertical line cannot be captured in the form \( y = mx + c \) since \( m \) is not defined. Such lines are of the form \( x = \text{constant} \). In our case, since the point \( (2,\ 3) \) lies on the tangent and its slope is undefined, the vertical tangent is:

\[ \boxed{x = 2}. \]

The other tangent, with slope \( \frac{5}{12} \), is:

\[ y = \frac{5}{12}x + 3 - 2 \cdot \frac{5}{12} = \frac{5}{12}x + 3 - \frac{10}{12} = \frac{5}{12}x + \frac{26}{12} = \frac{5}{12}x + \frac{13}{6}. \]

Hence, the two tangents from point \( (2,\ 3) \) to the circle \( x^2 + y^2 = 4 \) are:

\[ \boxed{x = 2} \quad \text{and} \quad \boxed{y = \frac{5}{12}x + \frac{13}{6}}. \]

This subtle case illustrates how vertical tangents arise when the general slope-based method appears to give only one solution. The geometry confirms the presence of two tangents.

Chord of Contact

Suppose from an external point \( P(x_1, y_1) \), two tangents are drawn to a circle, touching it at points \( Q \) and \( R \). Then the chord \( QR \), which joins the points of contact on the circle, is called the chord of contact of the point \( P \) with respect to the circle.

Let the equation of the circle be

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

with center at the origin and radius \( a \). Our aim is to find the equation of the chord of contact \( QR \).

A natural idea might be to:

1. Find the two tangents from \( P(x_1, y_1) \) to the circle.
2. Determine their respective points of contact \( Q(x_2, y_2) \) and \( R(x_3, y_3) \).
3. Use these two points to find the equation of the line joining them.

However, this approach is long and computationally intensive. Instead, we proceed more cleverly.

Let \( Q(x_2, y_2) \) and \( R(x_3, y_3) \) be the points where the tangents from \( P \) touch the circle. The tangent to the circle at any point \( (x, y) \) on the circle \( x^2 + y^2 = a^2 \) is:

\[ xx_0 + yy_0 = a^2, \]

where \( (x_0, y_0) \) is the point of contact.

So, the tangent at \( Q(x_2, y_2) \) is:

\[ xx_2 + yy_2 = a^2 \tag{1} \]

and the tangent at \( R(x_3, y_3) \) is:

\[ xx_3 + yy_3 = a^2. \tag{2} \]

Since both tangents pass through the external point \( P(x_1, y_1) \), point \( P \) must satisfy both (1) and (2). Hence:

\[ x_1x_2 + y_1y_2 = a^2 \tag{3} \]

\[ x_1x_3 + y_1y_3 = a^2. \tag{4} \]

Now consider the equation:

\[ xx_1 + yy_1 = a^2. \tag{5} \]

At first glance, this looks like the equation of the tangent to the circle at \( (x_1, y_1) \), but since \( (x_1, y_1) \) does not lie on the circle, this is not a tangent.

However, observe that when we substitute \( (x_2, y_2) \) and \( (x_3, y_3) \) into (5), both equations (3) and (4) imply that:

\[ x_2x_1 + y_2y_1 = a^2,\quad x_3x_1 + y_3y_1 = a^2. \]

So both \( Q \) and \( R \) lie on the line (5). Hence, equation (5) passes through both \( Q \) and \( R \), and therefore represents the chord of contact.

Thus, the equation of the chord of contact from point \( P(x_1, y_1) \) to the circle \( x^2 + y^2 = a^2 \) is:

\[ \boxed{xx_1 + yy_1 = a^2}. \]

This equation has the same form as that of the tangent to the circle, but is interpreted differently when \( (x_1, y_1) \) lies outside the circle.

General Case:

For a circle of the form

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

and an external point \( P(x_1, y_1) \), the equation of the chord of contact is:

\[ \boxed{xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0}. \]

This is identical in form to the equation of tangent to the circle at \( (x_1, y_1) \), except that now \( (x_1, y_1) \) does not lie on the circle.

In short, the chord of contact from point \( P(x_1, y_1) \) to the circle \( S \) is denoted compactly by the equation:

\[ \boxed{T = 0}, \]

where \( T \) represents the tangent form evaluated at \( (x_1, y_1) \).

Angle Between a Pair of Tangents

Let \( P(x_1, y_1) \) be a point outside a circle with center \( C \) and radius \( R \). From \( P \), draw two tangents \( PQ \) and \( PR \) to the circle, touching the circle at points \( Q \) and \( R \), respectively. Let us now compute the angle between these two tangents.

To understand the configuration, join the center \( C \) to \( Q \) and \( R \), and also join \( P \) to \( C \). Draw chord \( QR \) and drop a perpendicular from \( C \) onto it, intersecting at point \( N \). This construction is standard and arises frequently in probelms.

By symmetry and basic geometric reasoning, triangles \( \triangle PQC \) and \( \triangle PRC \) are congruent, since both share side \( PC \), and the segments \( CQ = CR = R \), and \( PQ = PR \) as tangents from the same external point. It follows that the angles \( \angle CPQ \) and \( \angle CPR \) are equal. Let each of these angles be \( \theta \). Then the total angle between the tangents is \( \angle QPR = 2\theta \).

Now consider triangle \( \triangle PQC \). It is a right-angled triangle at \( Q \), since the radius \( CQ \) is perpendicular to the tangent \( PQ \). We know:

• \( CQ = R \), the radius.

• \( PQ = \text{length of the tangent} = \sqrt{S_1} \), where

  \[ S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c, \]

when the general form of the circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \). For the standard circle \( x^2 + y^2 = R^2 \), we have:

\[ S_1 = x_1^2 + y_1^2 - R^2. \]

In triangle \( \triangle PQC \), we then have:

\[ \tan \theta = \frac{R}{\sqrt{S_1}}. \]

Using the identity:

\[ \sin(2\theta) = \frac{2\tan \theta}{1 + \tan^2 \theta}, \]

we substitute:

\[ \sin(2\theta) = \frac{2 \cdot \dfrac{R}{\sqrt{S_1}}}{1 + \left( \dfrac{R}{\sqrt{S_1}} \right)^2} = \frac{2R / \sqrt{S_1}}{1 + R^2/S_1} = \frac{2R\sqrt{S_1}}{R^2 + S_1}. \]

Hence, the angle between the two tangents is:

\[ \boxed{\angle QPR = 2\theta = \sin^{-1}\left( \frac{2R\sqrt{S_1}}{R^2 + S_1} \right)}. \]

This gives an exact expression for the angle between the pair of tangents drawn from a point \( P(x_1, y_1) \) to a circle of radius \( R \), using only coordinate geometry and trigonometric identities.

Length of Chord of Contact

Using the same geometric setup as in the previous section, let \( P(x_1, y_1) \) be a point outside a circle centered at \( C \) with radius \( R \). From \( P \), two tangents \( PQ \) and \( PR \) are drawn to the circle, touching it at points \( Q \) and \( R \), respectively. The line segment \( QR \) is the chord of contact.

We now aim to compute the length of this chord.

From the geometry, it can be seen that point \( N \) bisects \( QR \), due to symmetry. Triangle \( \triangle CNQ \) is a right-angled triangle at \( N \), with:

• \( \angle NQC = \theta \), the angle between the radius \( CQ \) and the tangent \( PQ \),
• \( QC = R \), the radius of the circle.

From elementary trigonometry:

\[ QN = QC \cdot \cos\theta = R \cos\theta. \]

Since \( QR = 2 \cdot QN \), the length of the chord of contact is:

\[ QR = 2R \cos\theta. \tag{1} \]

From the earlier calculation in the triangle \( \triangle PQC \), we had:

\[ \tan\theta = \frac{R}{\sqrt{S_1}}, \]

where \( S_1 \) is the value of the circle’s left-hand side evaluated at point \( P(x_1, y_1) \). Then,

\[ \cos\theta = \frac{1}{\sqrt{1 + \tan^2\theta}} = \frac{1}{\sqrt{1 + \frac{R^2}{S_1}}} = \frac{\sqrt{S_1}}{\sqrt{R^2 + S_1}}. \]

Substitute this into (1):

\[ QR = 2R \cdot \frac{\sqrt{S_1}}{\sqrt{R^2 + S_1}} = \boxed{ \frac{2R \sqrt{S_1}}{\sqrt{R^2 + S_1}} }. \]

This gives the exact length of the chord of contact drawn from point \( P(x_1, y_1) \) to a circle of radius \( R \).

Some Observations About the Geometry of Tangents from an External Point

Let \( P \) be an external point from which two tangents \( PQ \) and \( PR \) are drawn to a circle with center \( C \), and let \( Q \) and \( R \) be the respective points of contact. We have the following observations:

1. Cyclic Quadrilateral \( PQCR \):

Note that both \( \angle PQC \) and \( \angle PRC \) are right angles since the radius of a circle is always perpendicular to the tangent at the point of contact. Therefore,

\[ \angle PQC + \angle PRC = 90^\circ + 90^\circ = 180^\circ. \]

This implies that quadrilateral \( PQCR \) is cyclic. That is, its four vertices lie on a common circle—there exists a unique circumcircle passing through all four points.

2. P and C are Diametrically Opposite:

Since both \( \angle PQC \) and \( \angle PRC \) are right angles, and they subtend the same arc \( \widehat{QR} \), the line segment \( PC \) must be a diameter of the circumcircle of quadrilateral \( PQCR \). Hence, points \( P \) and \( C \) are diametrically opposite on this new circle.

Let us now write the equation of this circumcircle.

3. Equation of the Circumcircle of \( PQCR \):

If the points \( P \) and \( C \) are diametrically opposite on a circle, and the circle is centered at the midpoint of \( PC \), then the circle can be written using the diameter form of a circle equation. If \( P = (x_1, y_1) \) and \( C = (-g, -f) \), then the equation of a circle with diameter \( PC \) is:

\[ (x - x_1)(x + g) + (y - y_1)(y + f) = 0. \]

This equation can also be interpreted as the circle that passes through the four points \( P, Q, C, R \). In fact, the same circle is the circumcircle of all triangles formed by any three of the four points:

• Triangle \( \triangle PQR \),
• Triangle \( \triangle PQC \),
• Triangle \( \triangle PRC \),
• Triangle \( \triangle QRC \).

Director Circle

The director circle of a given circle is defined as the locus of a point from which two perpendicular tangents can be drawn to the given circle. That is, the angle between the pair of tangents from this point to the circle is \( 90^\circ \).

Let the given circle have radius \( r \) and be centered at point \( C \). Consider a point \( P \) outside the circle such that the two tangents drawn from \( P \) touch the circle at points \( Q \) and \( R \), and these tangents are perpendicular. Then:

• By definition, \( \angle QPR = 90^\circ \)
• Each tangent is perpendicular to the radius at the point of contact: \( \angle PQC = \angle PRC = 90^\circ \)
• This implies that triangle \( \triangle QCR \) is also right-angled at \( C \), and \( CQ = CR = r \)
• Hence, quadrilateral \( PQCR \) has all sides equal and all angles \( 90^\circ \), so it is a square
• Therefore, the point \( P \) lies on a circle of radius \( \sqrt{2}r \), centered at \( C \)

Thus, the locus of point \( P \) is a circle with:

• Same centre as the original circle
• Radius \( \sqrt{2}r \)

This circle is called the director circle.

For example, consider the circle:

\[ x^2 + y^2 = a^2 \]

The director circle has the same centre and radius \(\sqrt(2)a\) and its equation is:

\[ \boxed{x^2 + y^2 = 2a^2} \]