# Thread: Cirlce and line problem

1. ## Cirlce and line problem

new to cirlces and getting stumped at this question. I'll run through my workings and thoughts - any advise would be greatly received.

The equation of a cirlce with centre C is: $\displaystyle (x-3)^2+ (y-4)^2 = 9$ and O is the origin. the line joining O and C can be extended to meet the cirlce at P. Find the coordinates of P and show the equation of the tangent to the cirlce at P is $\displaystyle 3x + 4y = 40$.

So, I can observe the coords at C to be (3,4) and the line OP to have an equation of $\displaystyle y-0 = \frac{4}{3}(x-0)$ and therefore, $\displaystyle y=\fra{4}{3}x$.

Substitute this into the circle equation and I get $\displaystyle 75x^2 + 576x -108 = 0$ which solves to be approx. x = 0.1831 and -7.8631

But these x coords cant be right as I know the circle lies in the 1st quadrant and there for cannot have an intercept with the line at x = -7.8....

What has gone wrong? Once I get the coords for the point P I can calculate the new equation of the line.

2. Hello, dojo!

Your algebra went a little off . . .

$\displaystyle \text{The equation of a circle with centre }\,C\text{ is:}$
. . $\displaystyle (x-3)^2+ (y-4)^2 \:=\: 9\:\text{ and }O\text{ is the origin.}$

$\displaystyle \text{The line joining }O\text{ and }C\text{ is extended to meet the circle at }P.$

$\displaystyle \text{Find the coordinates of }P.$

$\displaystyle \text{Show the equation of the tangent to the circle at }P\text{ is: }3x + 4y \:=\: 40$

$\displaystyle \text{So, I can observe the coords at }C\text{ to be }(3,4)$ .Yes!
$\displaystyle \text{and the line }OP\text{ to have an equation: }\:y \:=\:\frac{4}{3}x$ .Right!

$\displaystyle \text{Substitute this into the circle equation}$
. . $\displaystyle \text{and I get: }\:75x^2 + 576x -108 \:=\: 0$ .no

Substitute: .$\displaystyle (x-3)^2 + \left(\frac{4}{3}x-4\right)^2 \:=\:9$

. . $\displaystyle x^2 - 6x + 9 + \frac{16}{9}x^2 - \frac{32}{3}x + 16 \:=\:0$

. . . . . . . . . . . . $\displaystyle \frac{25}{9}x^2 - \frac{50}{3}x + 16 \:=\:0$

Multiply by 9: .$\displaystyle 25x^2 - 150x + 144 \:=\:0$

Factor:. . . . . . $\displaystyle (5x - 6)(5x-24) \:=\:0$

. . Hence: .$\displaystyle x \;=\;\frac{6}{5},\:\frac{24}{5}$

Since $\displaystyle \,P$ is further from $\displaystyle \,O$. we have: .$\displaystyle \boxed{P\left(\tfrac{24}{5},\:\tfrac{32}{5}\right) }$

Line $\displaystyle \,OP$ has slope $\displaystyle \frac{4}{3}$

The tangent at $\displaystyle \,P$ is perpendicular to $\displaystyle OP.$
. . Hence, its slope is $\displaystyle -\frac{3}{4}$

The line through $\displaystyle \left(\frac{24}{5},\:\frac{32}{5}\right)$ with slope $\displaystyle -\frac{3}{4}$ is:

. . . . . . $\displaystyle y - \frac{32}{5} \:=\:-\frac{3}{4}\left(x - \frac{24}{5}\right)$

Simplify: .$\displaystyle y - \frac{32}{5} \:=\:-\frac{3}{4}x + \frac{18}{5}$

Multiply by 20: .$\displaystyle 20y - 128 \:=\:-15x + 72 \quad\Rightarrow\quad 15x + 20y \:=\:200$

Divide by 5: .$\displaystyle \boxed{3x + 4y \:=\:40}$

3. LOL. I was applying the substitution to another circle expression in the question above! Thanks for pointing this out...