Tangent and Normal at a point on the Circle

In the previous section, we found the equation of a tangent to a circle when the slope of the tangent was given. In this section, we determine the equation of the tangent to a circle at a given point lying on the circle. We also derive the equation of the normal, which is the line perpendicular to the tangent at that point and passes through it.

Tangent at a Given Point on a Circle

Consider a circle of the general form

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

whose center is \( C = (-g,\ -f) \). Let \( P(x_1,\ y_1) \) be a point on the circle. Then \( P \) must satisfy the circle’s equation:

\[ x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = 0. \tag{1} \]

The tangent at point \( P \) is perpendicular to the radius \( CP \), so we compute the slope of the radius \( CP \). Since the coordinates of the center are \( (-g,\ -f) \), the slope of \( CP \) is

\[ \frac{y_1 + f}{x_1 + g}. \]

Then, the slope of the tangent (being the negative reciprocal) is

\[ - \frac{x_1 + g}{y_1 + f}. \]

Thus, the equation of the tangent at \( P(x_1,\ y_1) \) is the line passing through \( (x_1,\ y_1) \) with this slope:

\[ \frac{y - y_1}{x - x_1} = -\frac{x_1 + g}{y_1 + f}. \]

Cross-multiplying gives:

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

Expanding both sides:

\[ yy_1 + fy - y_1^2 - fy_1 = -xx_1 - gx + x_1^2 + gx_1. \]

Rearrange as:

\[ xx_1 + yy_1 + gx + fy = x_1^2 + y_1^2 + gx_1 + fy_1 \]

Now, add \( gx_1 + fy_1 + c \) to both sides:

\[ xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c. \]

But from equation (1), the right-hand side is zero. Hence, the equation of the tangent at \( P(x_1,\ y_1) \) is:

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

This form is symmetric in \( (x,\ y) \) and \( (x_1,\ y_1) \), and directly expresses the tangent at a known point on the circle.

Example

Find the equation of the tangent to the circle

\[ x^2 + y^2 - 4x + 6y = 0 \]

at the origin \( (0,\ 0) \).

Solution:

We compare the given equation with the general form

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

to identify the parameters:

\[ 2g = -4 \implies g = -2,\quad 2f = 6 \implies f = 3,\quad c = 0. \]

We are to find the tangent at the point \( P(0,\ 0) \). First, we verify that this point lies on the circle:

\[ 0^2 + 0^2 - 4 \cdot 0 + 6 \cdot 0 = 0 \implies \text{Yes, origin lies on the circle}. \]

Using the formula for the tangent at \( (x_1,\ y_1) \), we write:

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

Substituting \( x_1 = 0, y_1 = 0, g = -2, f = 3, c = 0 \), the tangent equation becomes:

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

Thus, the equation of the tangent at the origin is

\[ \boxed{2x - 3y = 0}. \]

To quickly write the equation of the tangent to a circle at a given point \( (x_1, y_1) \) lying on it, consider the general form of the circle:

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

The equation of the tangent at \( (x_1, y_1) \) can be directly obtained by the following replacements:

• Replace \( x^2 \) with \( xx_1 \),
• Replace \( y^2 \) with \( yy_1 \),
• Replace \( x \) (in the linear term) with \( \frac{x + x_1}{2} \),
• Replace \( y \) (in the linear term) with \( \frac{y + y_1}{2} \),
• Keep the constant term \( c \) as it is.

Apply these substitutions to the original equation:

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

The tangent becomes:

\[ xx_1 + yy_1 + 2g \cdot \frac{x + x_1}{2} + 2f \cdot \frac{y + y_1}{2} + c = 0 \]

Simplifying:

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

This is exactly the general equation of the tangent to the circle at point \( (x_1, y_1) \).

Example

Find the equation of the tangent to the circle

\[ x^2 + y^2 + 6x - 8y + 5 = 0 \]

at the point \( (1,\ 2) \).

Apply substitution:

Replace \( x^2 \to x \cdot 1,\ y^2 \to y \cdot 2 \)
Replace \( x \to \frac{x + 1}{2},\ y \to \frac{y + 2}{2} \)
So the tangent is:

\[ x \cdot 1 + y \cdot 2 + 6 \cdot \frac{x + 1}{2} - 8 \cdot \frac{y + 2}{2} + 5 = 0 \]

Simplify:

\[ x + 2y + 3(x + 1) - 4(y + 2) + 5 = 0 \]

\[ x + 2y + 3x + 3 - 4y - 8 + 5 = 0 \]

\[ 4x - 2y = 0 \implies \boxed{2x - y = 0} \]

This is the required tangent at \( (1,\ 2) \).

Normal at a Point on the Circle

The normal to a circle at a given point is the line perpendicular to the tangent at that point. But more simply, since the radius of the circle is always perpendicular to the tangent and lies along the line from the center to the point, the normal is just the line that joins the center of the circle to the point on the circle.

Suppose the equation of the circle is

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

with center \( (-g,\ -f) \), and let \( (x_1,\ y_1) \) be a point on the circle.

Then the line passing through both \( (x_1,\ y_1) \) and the center \( (-g,\ -f) \) is the normal. Since we already know two points on it, the equation of the line is written using the section formula or directly using equal slopes of segments:

\[ \frac{y_1 + f}{x_1 + g} = \frac{y + f}{x + g}. \]

This is the required equation of the normal. There is no need to simplify it further or write it in any other form.

Calculation of Point of Contact of a Tangent with a Circle

When a line is tangent to a circle, we know it touches the circle at exactly one point. The condition of tangency tells us whether a line is tangent: for the standard circle

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

a line \( y = mx + c \) is a tangent if and only if

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

The question we now address is this:

If a given line is tangent to a circle, what is its point of contact with the circle?

There are multiple ways to find this point of contact.

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Method 1: Intersection of Tangent and Normal

Let the center of the circle be \( C \), and suppose the given tangent line \( L \) touches the circle at point \( P \). Since the radius \( CP \) is perpendicular to the tangent, its slope is the negative reciprocal of the slope of the tangent. If the tangent has slope \( m \), then the normal \( CP \) has slope \( -1/m \). As \( CP \) passes through the center \( C \), we can write its equation. The intersection of this normal with the tangent line gives the coordinates of the point of contact \( P \).

This method is clean and geometrically motivated.

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Method 2: Foot of the Perpendicular

From geometry, we know the point of contact \( P \) is the foot of the perpendicular from the center \( C \) onto the tangent line \( L \). The formula for the foot of perpendicular from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) gives the coordinates of \( P \) directly and efficiently.

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Method 3: Comparison Method (Algebraic)

This is an elegant algebraic method based on comparing the known tangent line with the general equation of the tangent to a circle at an unknown point.

Let the line be \( y = mx + c \), and suppose it is a tangent to the circle

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

so that the condition of tangency holds:

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

Let the point of contact be \( (x_1, y_1) \). Then, since \( (x_1, y_1) \) lies on the circle, the equation of the tangent at that point is:

\[ xx_1 + yy_1 = a^2. \tag{1} \]

The given line is:

\[ y = mx + c \quad \text{or} \quad -mx + y - c = 0. \tag{2} \]

Equations (1) and (2) represent the same line (since both describe the same tangent), so they must be proportional. That is, their coefficients must satisfy:

\[ \frac{x_1}{-m} = \frac{y_1}{1} = \frac{-a^2}{-c}. \]

Now solve:

\[ x_1 = -\frac{a^2 m}{c}, \quad y_1 = \frac{a^2}{c}. \]

Therefore, the point of contact of the tangent \( y = mx + c \) with the circle \( x^2 + y^2 = a^2 \) is:

\[ \boxed{\left( -\frac{a^2 m}{c},\ \frac{a^2}{c} \right)}. \]

This method is purely algebraic and uses the known form of the tangent at a point on the circle to force the point of contact by comparing coefficients.

Example

The line \( y = x + 1 \) cuts the circle

\[ x^2 + y^2 = 5 \]

at two points \( A \) and \( B \). The tangents to the circle at points \( A \) and \( B \) intersect at a point \( P \). Find the coordinates of point \( P \).

Solution:

Let the point of intersection of the two tangents, i.e., point \( P \), be \( (x_1, y_1) \). Then, from the geometry, the line \( AB \), which joins the points of contact, is the chord of contact from point \( P \) to the circle.

Therefore, the line \( y = x + 1 \) must be the chord of contact with respect to point \( P \). The chord of contact to the circle \( x^2 + y^2 = 5 \) from point \( (x_1, y_1) \) has the equation:

\[ xx_1 + yy_1 = 5. \tag{1} \]

But we are given that the chord is also the line:

\[ y = x + 1 \implies x - y + 1 = 0. \tag{2} \]

Now, equations (1) and (2) both represent the same straight line, so their coefficients must be proportional. Write equation (1) in the same form:

\[ xx_1 + yy_1 - 5 = 0. \tag{3} \]

Now match the coefficients from (2) and (3):

\[ \frac{x_1}{1} = \frac{y_1}{-1} = \frac{-5}{1}. \]

From this, we obtain:

\[ x_1 = -5,\quad y_1 = 5. \]

Hence, the point \( P \), the intersection point of the two tangents drawn at \( A \) and \( B \), is:

\[ \boxed{(-5,\ 5)}. \]