Circle geometry once more!

**booper563** · December 12th 2008, 08:10 AM

A circle with the centre (-1,-2) is tangent to the line 3x+4y-14=0

Determine the radius of the circle algebraically.

The part that confuses me is the equation there, I'm not sure what to do with it or anything.
so please give me step by step details, they would be greatly appreciated.

**running-gag** · December 12th 2008, 08:21 AM

Hi

If you know the formula of the distance between a point and a line it is very easy since the radius of the circle is the distance between the center C and the tangent line (D)

The formula for a point A(xA,yA) and a line (d) ax+by+c=0 is

$d(A,(d)) = \frac{|a x_A + b y_A + c|}{\sqrt{a^2 + b^2}}$

$R = d(C,(D)) = \frac{|3 x_C + 4 y_C - 14|}{\sqrt{3^2 + 4^2}} = 5$

If you don't know the formula then you can use the point of tangence H with the conditions
(i) H is on the line (D)
(ii) (CH) is perpendicular to (D)

**booper563** · December 12th 2008, 08:29 AM

Is the formula d= square root of (x2-x1)^2 +(y2-y)^2

**running-gag** · December 12th 2008, 08:33 AM

The formula you have written is the distance between 2 points M1(x1,y1) and M2(x2,y2)
$M_1M_2 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

**booper563** · December 12th 2008, 08:35 AM

So for that formula, what numbers to i plug into it, im so confused,

**running-gag** · December 12th 2008, 09:05 AM

If you are not allowed to use the formula I gave you in my first post, here is how you can manage
Let H be the tangence point of the circle and the line
Then
(i) $\overrightarrow{CH}.\overrightarrow{u}=0$ where u is a direction vector of the line
(ii) 3xH+4yH-14=0

This gives 2 equations, the resolution gives the coordinates of H

Tell me if you need additional help

**booper563** · December 12th 2008, 09:10 AM

no no, im allowed using the formula.
I'm just really confused, I dunno were to put numbners and what im suppose to end up with

**running-gag** · December 12th 2008, 09:19 AM

The distance between a point A(xA,yA) and a line (d) ax+by+c=0 is

$d(A,(d)) = \frac{|a x_A + b y_A + c|}{\sqrt{a^2 + b^2}}$

You are looking for the distance betwwen the center C(-1,-2) and the line 3x+4y-14=0
The formula gives

$d = \frac{|3 \cdot (-1) + 4 \cdot (-2) - 14|}{\sqrt{3^2 + 4^2}} = \frac{|-25|}{\sqrt{25}} = 5$

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