1. ## Hard geometry question

Hey guys,
just been revising and I have came across this question but don't know how to solve it.

This question is part d of the question (with 3 sub-parts to it)

I have already found the centre of the circle and the radius on the previous parts of the question. Center (-7,5), r=5 but dont know if this comes in use to this part of the question.

The question being

A line has the equation y=kx+6, where k is a constant.

i) show that x-coordinates of any point of intersection of the line and the circle satisfy this equation (k^2+1)x^2+2(k+7)x+25=0.

ii) The equation (the one above) has equal roots. Show that 12k^2-7k-12=0

iii) Hence find the values of k for which the line is a tangent to the circle

Danthemaths

2. ## Re: Hard geometry question

Graph the circle and given point (0,6).Find the lengths of the two equal tangents and then the equations of each tangent line.Start with the distance from center to (0,6)

3. ## Re: Hard geometry question

where did you get the center to be (0,6)?

4. ## Re: Hard geometry question

Originally Posted by Danthemaths
where did you get the center to be (0,6)?
the center is ( -7,5), (0,6) is the y intersept of y=kx +6

5. ## Re: Hard geometry question

The attached pdf may help you. If that equation "has equal roots," that means the disciminant b^s -4ac must be equal to zero.

The two tangents are not "equal", but they are perpendicular. There are two unequal values of k for which that equation has equal roots. Perhaps you were looking for equal values of k and that's what confused you? Just guessing.

The point of the exercise seems to be that through any point external to the circle, two tangents can be constructed.

6. ## Re: Hard geometry question

Originally Posted by zhandele
The attached pdf may help you. If that equation "has equal roots," that means the disciminant b^s -4ac must be equal to zero.

The two tangents are not "equal", but they are perpendicular. There are two unequal values of k for which that equation has equal roots. Perhaps you were looking for equal values of k and that's what confused you? Just guessing.

The point of the exercise seems to be that through any point external to the circle, two tangents can be constructed.
Tangents to a circle from an external point are comgruent. These tangents can be constucted by finding the length of one using thr Pythagoran Theorem and then a compass adjusted to that length.To locate the precise point of tangency requires finding the slope of the tangent