The equation of a cirlce is (x+1)(squared) + y(squared)=13
i)Write down the centre and radius of this cirlce
ii)find the equation of the tangent to this cirlce at the point (2,2)
Thank you
$\displaystyle (x - h)^2 + (y - k)^2 = r^2$
is the general equation of a circle where the center is at (h, k) and the radius is r.
As to the tangent line, the first derivative is the slope of the tangent line, right? so take the derivative (implicitly):
$\displaystyle (x + 1)^2 + y^2 = 13$
$\displaystyle 2(x + 1) + 2y \frac{dy}{dx} = 0$
$\displaystyle \frac{dy}{dx} = -\frac{x + 1}{y}$
So at (2, 2) the slope of the tangent line is
$\displaystyle \frac{dy}{dx} = -\frac{3}{2}$
So find the line with a slope of -3/2 that passes through the point (2, 2).
-Dan
Centre: (-1 ; 0)
Radius: $\displaystyle \sqrt{13}$
The tangent must make an angle of 90 degrees to the radius. (Sorry i forgot the proper English word for that)
So let's work out the equation from the radius to that (2;2) point.
$\displaystyle y_{r} = mx + c$
$\displaystyle = \frac{2}{3} x + c$
Solve for c and you'll find:
$\displaystyle y_{r} = \frac{2}{3} x + \frac{2}{3}$
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Now we must find a line that has a gradient of $\displaystyle \frac{-3}{2}$
$\displaystyle y_{t} = mx + c$
$\displaystyle = \frac{-3}{2} x + c$
Solve for c by substituting the points (2 ; 2)
[Note: You cannot use (-1;0)!! The line does not pass through that point.]
$\displaystyle y_{t} = \frac{-3}{2} x + 5$