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1. line PQ is a diameter of circle centre (-4,-2).line passes through P and is perpendicular to PQ. given Q is (10.4) find the equation of L.
2. The circle's centre C is the midpoint of PQ.

CQ = (10, 4) - (-4, -2) = (14, 6)
CP = -(14, 6)
P = (-4, -2) - (14, 6) = (-18, -8)

Gradient of PQ = 6/14 = 3/7

Equation of L:
y = -(7/3)(x + 18) - 8
y = -(7/3)x - 50
3. no offence but could u explain what went on there ??? andf thanks
4. To go from C to Q, you have to move 14 units right and 6 up. You have to make the opposite movement to go from C to P: 14 units left and 6 down. So (coordinates of P) = (coordinates of C) - (14, 6) = (-18, -8).

The gradient of PQ is 6/14 = 3/7. The gradients of perpendicular lines have a product of -1. So the gradient of L is -(7/3).

Then you have to find the equation of the line of gradient -(7/3) that passes through (-18, -8).
5. If u ve drawn a diagram notice that the line is tangent to the circle at P, Q(10,4) is at the other far end of the circle and C(-4,-2) is the centre.

As C is the centre of the circle it is the midpint of the diameter PQ.

Let the coordinates of P(x,y)

By the equation of the midpoint u get:
(-4,-2)=[(10+x)/2],[(4+y)/2]

By Equating u u get the coordinates of P (-18,-8)
Find the gradient of the line PQ which is equal to 3/7

The required line is Perp to PQ hose gradient is (-7/3)

Plug the values in to find the value of the equation

I hope u understand
6. thanks buddies great stuff

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