Radius & Tangent (Edexcel A-Level Mathematics): Revision Notes

• Tangent: A tangent is a line that just touches a circle but does not cut it.
• Normal (or radius): A normal is at right angles to a tangent.

Key Properties of a Radius:

1. All radii of a circle are equal in length.
2. A radius is always perpendicular to the tangent at the point of contact.

Key Properties of Tangents:

1. A tangent touches the circle at only one point.
2. A tangent to a circle is always perpendicular to the radius at the point of tangency.
3. From a point outside the circle, you can draw two tangents to the circle. These tangents are equal in length.

For a circle with equation $x^2 + y^2 = 36$:

• At point $A(6, 0)$, the tangent would be:

This is a vertical line, tangent to the circle at $(6, 0)$.

• At point $B(0, 6)$, the tangent would be:

This is a horizontal line, tangent to the circle at $(0, 6)$.

Example: Find the equation of the tangent to the circle

$x^2 + y^2 + 4x - 13 = 0$

at the point $(-1, 4)$.

1. Rewrite the circle equation: $(x + 2)^2 + y^2 = 17$

• Completing the square for $x$reveals the centre of the circle: $(-2, 0)$.

4. Find the centre:

• The centre of the circle is $(-2, 0)$.

5. Find the gradient of the normal: $m = \frac{4 - 0}{-1 + 2} = 4$

• Gradient of the normal is $4$.

6. Find the gradient of the tangent: $\text{Gradient of tangent} = -\frac{1}{4}$

• The gradient of the tangent is the negative reciprocal of the gradient of the normal.

7. Equation of the tangent: $y - 4 = -\frac{1}{4}(x+1)$

• Using the point-slope form of the line equation.

8. Simplify the equation: $y - 4 = -\frac{1}{4}(x + 1)$

$4y - 16 = -(x + 1)$

$4y - 16 = -x - 1$

$x + 4y - 15 = 0$

Final Equation

$x + 4y - 15 = 0$

This is the equation of the tangent to the circle at the point $(-1, 4)$.

Problem:

The line with the equation $y = x + k$ is a tangent to the circle with the equation $x^2 + y^2 + 6x - 8y + 17 = 0$. Find the two possible values of $k$.

Solution:

1. Substitute $y = x + k$ into the circle equation: $x^2 + (x + k)^2 + 6x - 8(x + k) + 17 = 0$

2. Expand and simplify: $x^2 + x^2 + 2kx + k^2 + 6x - 8x - 8k + 17 = 0$

$2x^2 + 2kx - 2x + k^2 - 8k + 17 = 0$

3. Group the terms: $2x^2 + (2k - 2)x + (k^2 - 8k + 17) = 0$

4. Identify coefficients for the quadratic equation in terms of $x$: $a = 2$

$b = 2k - 2$

$c = k^2 - 8k + 17$

5. Since the line is a tangent, the quadratic equation has exactly one solution:

• This implies the discriminant $\Delta$ must be zero. $\Delta = b^2 - 4ac = 0$

6. Substitute the values of $a, b$, and $c$ into the discriminant formula: $(2k - 2)^2 - 4(2)(k^2 - 8k + 17) = 0$

$(2k - 2)^2 - 8(k^2 - 8k + 17) = 0$

$4k^2 - 8k + 4 - 8k^2 + 64k - 136 = 0$

$-4k^2 + 56k - 132 = 0$

7. Solve the quadratic equation: $4k^2 - 56k + 132 = 0$

$k^2 - 14k + 33 = 0$

8. Factorise or use the quadratic formula: $k = \frac{14 \pm \sqrt{14^2 - 4 \cdot 33}}{2}$

$k = \frac{14 \pm \sqrt{64}}{2}$

$k = \frac{14 \pm 8}{2}$

$k = 11 \quad \text{or} \quad k = 3$

Final Values of $k$

$k = 11 \quad \text{or} \quad k = 3$