So let us talk about tangent of a circle. Now tangent is always the same. It doesn't matter what it is a tangent of. It never changes.
Tangent is a straight line that touches a curve at a point. As far as touching a curve at a point and is a straight line, that's the definition of a tangent, right? So To find your tangent of the circle, a couple of things you need to know. The first thing was talked about in Euclidean geometry. Now, we learned in Euclidean geometry that whenever you have a radius, so let's say that this here is O.
and this here is a we say that if OA is a radius the tangent is always at 90 degrees to your radius yeah so we always say our radius is perpendicular to a tangent now this is one thing you always need to put at the back of your mind if you remember that the radius is always perpendicular to the tangent We bring you back to analytical geometry and in analytical geometry we said when we have two lines that are perpendicular the gradient multiply to give you negative one. What I mean by that is that if I say that the line of our radius we call this gradient one and our tangent line we call this gradient two. We understand that if I get gradient one times gradient two it's going to give me negative one right.
Because whenever it's at 90 degrees or those two lines are perpendicular, we expect the product of your gradient to be negative 1. The first thing I said about a tangent is that a tangent is a straight line. And what we need to understand is that the equation of a straight line looks like this. Where you have y is equal to mx plus c, right?
You could also rewrite it like this, depending on what we're given. y minus y1 is equal to m open bracket x minus x1. and you need to understand that the gradient i'm going to substitute here this m right would be the m2 that you calculate from this one over here so if you want to do some calculations you understand that m2 would actually be negative 1 over m1 so if i'm able to find my value for m2 i just need to substitute the gradient into the formula right And after substituting the gradient, I just need to find the point, right? Any point on this line, right?
As far as I can get a point on the line, I substitute those two points as my x and y, and I'll be able to get my equation of a tangent. Now, that's exactly how you find your equation of any tangent to a circle. Now, this whole thing might be confusing. So let's do an example to try to put some of this into perspective.
Now this question says find the equation of the tangent APB which touches a circle. Center C with equation x minus 3 squared plus y plus 1 squared which is equals to 20. p is 5 and 3. Okay. Now, the question is asking us to find the equation of a tangent.
And a tangent is a straight line. Every straight line would always look like this. y is equal to mx plus c. And one of the important things in finding a straight line is figuring out what the gradient is. So I need to figure out my gradient and figuring out the gradient would definitely help me find my answers.
Before we get into that, a couple of things we notice in the question. The question here tells us that our center, that is C. Using those values, we understand that the coordinate at the center is actually positive 3 and negative 1. Now, from Euclidean, we understand that the radius and the tangent are usually perpendicular. So, if this is M1, that is M2.
Now, because they are perpendicular, we understand that the product of our gradient is negative 1. So, in this case here, I can just write it. as negative 1 over m1. So I need to figure out the value for my gradient then substitute it into the equation of our straight line. Using gradient formula, I can calculate the gradient of PC which would be So the gradient of PC is actually 2. Now substituting this 2 back into this, we understand that the gradient of M2 would be negative 1 over 2. That would be the gradient of our tangent.
In this case here, remember M2 is our AB, right? So having the gradient of eb, we can substitute this into this formula, which will help us find our equation of our tangent. Now, the gradient here is negative 1 over 2. Now, we just need a coordinate on our line. And since we're given p here as 5 and 3, we can substitute 5 and 3 into this formula.
So through simplification, we notice that the equation of our tangent is y is equals to negative 1 over 2x plus 11 over 2. And that's it. So this is exactly how you can calculate your equation of a tangent. Yeah, I hope this makes sense.