This lesson will cover a few examples, illustrating equations of tangents to circles, and their points of contacts.

**Example 1 **Find the equation of the tangent(s) of slope 4/3 to the circle x^{2} + y^{2} = 25. Also find the point(s) of contact.

**Solution **This problem is a direct application of the slope form of the tangent: \( y = mx ± a \sqrt{1+m^2} \) . The required equation is \(y = 4x/3 ± 5 \sqrt{1+(4/3)^2} \) , or **3y = 4x ± 25**.

To find the points of contact, we’ll use the coordinates derived in the previous lesson: (–a^{2}m/c, a^{2}/c). The required points are **(4, -3)** and **(-4, 3)** (Here m = 4/3 and c = ± 25/3). Have a look at the following figure for clarity

**Example 2 **Find the equation of the tangent(s) of slope 4/3 to the circle x^{2} + y^{2} – 4x – 6y + 12 = 0. Also find the point(s) of contact.

**Solution **Another problem involving direct application of the formula. But in this case, the center of the circle is not at the origin. We’ll use the second equation: \( (y + f) = m(x + g) ± r\sqrt{1+m^2} \)

The required equation is \( (y – 3) = 4(x – 2)/3 ± 1 \sqrt{1+(4/3)^2} \) . Or, **4x – 3y + 6 = 0** and **4x – 3y – 4 = 0**.

To find the points of contact, we’ll apply the formula again. I’ll skip the calculations. The required points are **(6/5, 18/5)** and **(14/5, 12/5)**.

Psst! There’s another way to find the point of contact – it will be the foot of perpendicular from the center to the tangent.

**Example 3 **Find the value(s) of c for which y = 2x + c is a tangent to the circle x^{2} + y^{2} = 1

**Solution **We’ve already done a similar problem earlier, using condition of tangency. Here, I’ll show a different approach.

We know that the line y = mx + c will be a tangent to the circle x^{2} + y^{2} = a^{2} when c = ±a\( \sqrt{1+m^2} \) . Here a = 1, and m = 2, therefore c = ±1\( \sqrt{1+2^2} \) or ± \( \sqrt{5} \).

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**Example 4 **Find the value(s) of m for which 3x + 4y = c is a tangent to the circle x^{2} + y^{2} – 4x – 6y – 12 = 0.

**Solution **Again, I’ll use the formula instead of condition of tangency, just to illustrate. Any tangent to the given circle looks like \( (y – 3) = m(x – 2) ± 5 \sqrt{1+m^2} \) or \( y = mx + 3 – 2m ± 1 \sqrt{1+m^2} \) .

The slope of the given line is –3/4. So all we have to plug in this value in the slope form of the tangent. We get y = –3x/4 + 3 – 2(–3/4) ± 5\( \sqrt{1+(3/4)^2} \) or 3x + 4y = 43, and 3x + 4y = –7. Therefore, c = 43 or – 7.

That’ll be all for this lesson. Next, I’ll be talking about tangents **at** a point. See you there!

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