the origin O is the midpoint of a chord PQ of the circle $\displaystyle (x - 6)^2 + (y + 3)^2 = 65$, show the gradient of the chord PQ is 2
could someone please show me how you would figure this out, thanks
Equation of chord of a circle when its midpoint is given is $\displaystyle S_1=S_11$ where S is the equation the circle.
Given equation is
$\displaystyle
x^2 + y^2 - 12x +6y -20 = 0$
Equation of a chord when its midpoint $\displaystyle (x_1,y_1)$
is $\displaystyle xx_1 + yy_1 - 6(x+x_1) + 3(y+y_1) - 20 = x_1^2 + y_1^2 - 12x_1 + 6y_1 -20 $
Here midpoint is 0rigin
Therefore
$\displaystyle x(0) + y(0) - 6(x+0) + 3(y +0) - 20 = 0^2 + 0^2 -12(0) + 6(0) - 20 $
Therefore equation of chord is
$\displaystyle 6x - 3y = 0 $ whose slope is -6/-3 =2
Okay I will suggest another method.
Let
y= mx be chord to circle
centre is C (6,-3)
Mid pt is T(0,0)
Slope of CT = 0-(-3)/0-6 = -1/2
Since CT is perpendicular to the chord, slope of the chord = m= 2
the equation becomes y = 2x
therefore gradient = 2
i think there's something wrong here, the book asked find the length of the chord PQ and the answer was $\displaystyle 4\sqrt{5}$ but the length of the centre to the origin is $\displaystyle 3\sqrt{5}$ which i think is what you were talking about. the centre (6, -3) is labelled as C by the way, so i'm not sure if it has anything to do with PQ
Well, I made a mistake by taking the midpoint (0,0) to be P randomly. I've changed the midpoint now to T. I didn't notice that the ends of the chord were P and Q. So let T (0,0) be the midpoint of PQ. Now the length of CT is $\displaystyle 3\sqrt5$. But $\displaystyle \Delta$ CTP is a right angled triangle. therefore,
By pythogarus theorem
$\displaystyle
CP^2 = CT^2 + TP^2 $
where TP = 1/2 length of chord.
Also CP = radius = $\displaystyle \sqrt65$
$\displaystyle \sqrt65^2 = 3\sqrt5^2 + TP^2$
$\displaystyle TP^2 = 20 $
$\displaystyle TP = 2\sqrt5$
Length of chord = 2 x TP = $\displaystyle 4\sqrt5$