# Tangent from an external point: Examples

This lesson will cover a few examples relating to tangents to a circle from an external point, which was covered in the last three lessons. Let’s begin.

Example 1 Find the equation of the tangents to the circle x2 + y2 = 25, from the point (7, 1).

Solution To solve this one, I’ll use the first method to solve this problem, as discussed in the lesson.

Any tangent to the circle x2 + y2 = 25 looks like y = mx + 5$$\sqrt{1+m^2}$$ .

Now this tangent is drawn from the point (7, 1). So we’ll simply substitute the coordinates of the given point in the tangent, and find the values of m.

On substitution, we get 1 = 7m + 5$$\sqrt{1+m^2}$$  or (1 – 7m)2 = 25(1 + m2).

Solving this equation for ‘m’ gives us two values m = 4/3 and -3/4, which makes sense, as there should be two different tangents from an external point.

Now as I told you before, we shouldn’t substitute these values of m back in the slope form of the tangent, because this will give us four equations of tangents (two coming from y = mx + 5$$\sqrt{1+m^2}$$  and the other two from mx – 5$$\sqrt{1+m^2}$$ ).

Instead, we’ll use the point form.

The required equations will be (y – 1) = (4/3)(x – 7) and (y – 1)=(-3/4)(x – 7), or 4x – 3y – 25 = 0 and 3x + 4y – 25 = 0.

Example 2 Find the equation of the tangents to the circle x2 + y2 – 6x – 8y = 0, from the point (2, 11).

Solution This time, I’ll use the second method, that is the condition of tangency, which is fundamentally same as the previous method, but only looks a bit different.

Any line through the given point is (y – 11) = m(x – 2), or mx – y + 11 – 2m = 0.

Now, for this line to be a tangent to the given circle, it’s distance from the center of the circle must be equal to it’s radius.

Applying the formula, we get |m + 7|/$$\sqrt{1+m^2}$$  = 5. Or, m2 + 14m + 49 = 25 + 25m2. Or, 12m2 – 7m – 12 = 0.

This gives us the values of m as 4/3 and -3/4.

Therefore the required tangents are y – 11 = (4/3)(x – 2) and (y – 11) = (-3/4)(x – 2), or 4x – 3y + 25 = 0 and 3x + 4y – 50 = 0

Example 3 Find the angle between the tangents to the circle x2 + y2 = 25, form the point (6, 8).

Solution I’ll use the slope form of the equation in this example to find the angle between the tangents, as discussed in this lesson.

Any tangent to the circle will be y = mx + 5$$\sqrt{1+m^2}$$

Substituting the point’s coordinates in the equation gives us 8 = 6m + 5$$\sqrt{1+m^2}$$

On rearranging, squaring and rearranging we get 11m2 – 96m + 39 = 0.

Remember we don’t need to solve the above equation. All we need are the values of |m1 – m2| and m1m2, which we’ll plug in the formula for angle between two lines.

Now |m1 – m2| = 50$$\sqrt{3}$$/11 and m1m2 = 39/11. Therefore, tanθ = |m1 – m2|/(1 + m1m2) = $$\sqrt{3}$$ which means the angle between the tangents is 60°.

Example 4 Find the angle between the tangents to the circle x2 + y2 – 4x – 2y – 20 = 0 form the point (7, 6)

Solution This one is similar to the previous one. But I’ll use geometry this time. The center of the circle is (2, 1) and the radius is 5.

The distance between the center and the given point (7, 6), using the distance formula, is $$\sqrt{50}$$  or 5$$\sqrt{2}$$.

So, if α be the angle between the tangents, then sin(α/2) = 5/5$$\sqrt{2}$$  or 1/$$\sqrt{2}$$ . This gives us α/2 as 45° or α = 90°

Example 5 Find the length of the tangent to the circle x2 + y2 = 12, drawn from the point (5, 6)

Solution This one is easy. Recall that to find the length of the tangent, all we have to do is substitute the coordinates of the point in the equation of the circle and take it’s square root.

The equation of the circle is x2 + y2 – 12 = 0. The required length will be  $$\sqrt{5^2 + 6^2 – 12}$$ = 7

And that’ll be all about tangents! Hope you’ve enjoyed the lessons.

The techniques used here will be used again in the forthcoming chapters, i.e. Parabola, Ellipse and Hyperbola. So make you’re through with these concepts.

We’ll now move on to something known as a Normal. See you in the next lesson.