C1!

What's the equation of the line that goes through O and C?

$(x-4)^{2} + (y-3)^{2} = 25$ that's the equation for centre of the circle

both co-ordinates are negative, so not sure how they got that!!

That didn't answer my question, I asked for the equation of the LINE that goes through O and C.



Drawing a graph makes this very clear.

equation of line is 3x+4y = 24

That is the equation of the line that goes through the points A, B and C. It doesn't go through the origin. You need to work out the equation of the line that goes through the origin and the point C. You have two coordinates, the general formula for the equation of a line when you know the coordinates of two points is

$\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}$ where the two points are $(x_1, y_1)$ and $(x_2, y_2)$

Have you drawn a graph?

I am just getting more confused here tbh..... equation of the line through O and C where the co-ordinates are O(0,0) and C(4,3)----- so gradient would just be 3/4

so we have y - 0 = 3/4(x) - 0 so y = 3/4(x)???

I am being very clear. Yes, the equation of the line that goes through O and C is indeed $y = \frac{3}{4}x$.
$\dfrac{y - 0}{3 - 0} = \dfrac{x - 0}{4 - 0}$

$4y = 3x$ so $y = \frac{3}{4}x$.

We want to know the coordinates when this line intersects with the tangent of the circle at B.

What's the equation of the tangent to the circle at B. We already know the equation of the normal at B, how can we use this to find the gradient of the tangent?

All i know is the co-ordinates of b which is (0,6)..... do we know the equation of the normal at B???? sorry for wasting your time like this mate!!

The normal to the tangent at B goes through the centre of the circle and the point B. It's fine, I don't mind helping you as long you try your best to help yourself, if that makes sense. Randomly spouting the equation of a circle will not get you to the answer.

Alright, let's start from the beginning so i know what i understand. It says, "the tangent to circle at B...." so the tangent is at the point (0,6) now when there is a tangent, there has to be a repeated root because two points are co-inciding, therefore can we sub in y = 6 into the equation of the circle???

I'm going to put all the existing information here in the order that you should approach the question.

Line 3x+4y = 24 is the line that goes through A, B and C (centre of circle), otherwise known as the normal to the tangent of the circle at B.

Rearrange this equation. $y = \frac{-3}{4}x + 6$

The gradient $m_1$ of this line is $\frac{-3}{4}x$

We want the gradient of the tangent. The gradient of a tangent is the reciprocal of the gradient of its normal where $m_1 \times m_2 = -1$

This makes the gradient of the tangent at B $\frac{4}{3}$.

What's the equation of the tangent now you have the gradient and a set of coordinates. We want to know when this line intersects the line that goes through O and C.

We know that the equation of that line is $y = \frac{3}{4} x$

They meet at the point D. How do we find the point where two equations meet?

Oh dear me, kill me, kill me now...... I got caught up in the wording, i missed the basic math ...... gradient is 4/3, we have the co-ordinates of (0,6) from that, we can get the equation...... therefore to intersect we put the equal to each other and solve..... that's all i needed to do and took me a while lol...... you're basically gave me the answer and i am very thankful mate..... i will +rep you tomorrow and rep you again for your patience towards me lol... thanks once again mate!!!