chord

• Oct 18th 2009, 06:20 AM
mark
chord
the origin O is the midpoint of a chord PQ of the circle $(x - 6)^2 + (y + 3)^2 = 65$, show the gradient of the chord PQ is 2

could someone please show me how you would figure this out, thanks
• Oct 18th 2009, 06:29 AM
nolanfan
Equation of chord of a circle when its midpoint is given is $S_1=S_11$ where S is the equation the circle.

Given equation is
$
x^2 + y^2 - 12x +6y -20 = 0$

Equation of a chord when its midpoint $(x_1,y_1)$
is $xx_1 + yy_1 - 6(x+x_1) + 3(y+y_1) - 20 = x_1^2 + y_1^2 - 12x_1 + 6y_1 -20$
Here midpoint is 0rigin
Therefore
$x(0) + y(0) - 6(x+0) + 3(y +0) - 20 = 0^2 + 0^2 -12(0) + 6(0) - 20$

Therefore equation of chord is
$6x - 3y = 0$ whose slope is -6/-3 =2
• Oct 18th 2009, 06:32 AM
mark
i don't think i understand that, is that called differentiation? because i haven't started doing that yet. is there a more simple way to do it?
• Oct 18th 2009, 06:56 AM
nolanfan
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

• Oct 18th 2009, 07:16 AM
mark
i think there's something wrong here, the book asked find the length of the chord PQ and the answer was $4\sqrt{5}$ but the length of the centre to the origin is $3\sqrt{5}$ 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
• Oct 18th 2009, 07:56 AM
nolanfan
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 $3\sqrt5$. But $\Delta$ CTP is a right angled triangle. therefore,
By pythogarus theorem
$
CP^2 = CT^2 + TP^2$

where TP = 1/2 length of chord.

Also CP = radius = $\sqrt65$

$\sqrt65^2 = 3\sqrt5^2 + TP^2$

$TP^2 = 20$

$TP = 2\sqrt5$

Length of chord = 2 x TP = $4\sqrt5$