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A level maths question

16 A circle has the centre (5, 3), and the y axis is a tangent to the circle.
(i) Write down the radius of the circle. [1 mark
(il) Write down the equation of the circle.
[2 mark
(ill) Show that the circle intersects the x axis at (1, 0). [2 mark
(iv) Find the points on the circle where the tangent is parallel to the diameter through (1, 0).
[4 mark
I have done the first 3 parts, I’m stuck on the last part which is 4 marks.
Original post by KindPot
16 A circle has the centre (5, 3), and the y axis is a tangent to the circle.
(i) Write down the radius of the circle. [1 mark
(il) Write down the equation of the circle.
[2 mark
(ill) Show that the circle intersects the x axis at (1, 0). [2 mark
(iv) Find the points on the circle where the tangent is parallel to the diameter through (1, 0).
[4 mark
I have done the first 3 parts, I’m stuck on the last part which is 4 marks.


You should be able to get the gradient of the diameter, then use that to find where the two tangents with the same gradient intersect the circle? Or think about the diameter which is normal to the two tangents.
Though a sketch and a bit of simple geometry should give the solns as well.
(edited 1 year ago)
Reply 2
393C5C91-93B0-4C2E-8AC2-BDD234E59C4F.jpeg
my working out so far
(edited 1 year ago)
Original post by KindPot
393C5C91-93B0-4C2E-8AC2-BDD234E59C4F.jpeg
my working out so far

ok, looks a bit complex. You know c as the line passes through the center though it seems to disappear in your working 1/2 way through and Im not sure why and the working looks incorrect from that point on.

A reasonable sketch and thinking about how the point (1,0) is transformed to the two points of interest, doing some very simple geometry if necessary is really the simplest way to go. For 4 marks, they probably want you to do this, even though you can grind the algebra as per the previous post/your attempt. Its obviously related to a 3 : 4 : 5 right triangle as per the previous parts of the question.

Edit - you could even do simple trig which is equivalent to the geometrical approach. Sometimes its easy to overlook the simple approaches and dive into the line intersecting circle algebra.
(edited 1 year ago)

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