1. ## Calculating the coordinates on the circumference using the radius

Hello friends,

I'm new on this forum. I'd be very grateful if you could please advise on how to solve this problem: just the approach would be greatly appreciated please - I've been stuck on this for days - I'd be grateful if you could please help me.

I need to find the coordinates on the circumference whereby the radius of the circle bisects a chord and touches that particular point. I've attached two images illustrating the situation.

I've solved the first three parts without any problems, but the last one is really bugging me. The question is actually one of the challenging ones on the exercise:

4. a) Show that the points P(2, -1) and Q(6, 7) lie on the circle whose equation is:

$x^2 + y^2 - 12x - 4y + 15$

(again, because I'm new, in case that equation doesn't appear, the equation is x^2 + y^2 - 12x - 4y + 15).

I've done this by subbing P and Q into that equation. PQ therefore form a chord.

b) Find the coordinates of the point M which is the midpoint of the chord PQ.

Midpoint is (4, 3).

c) Find the equation of the radius of the circle that passes through M.

The equation is $y = 5 - 1/2x$ (y = 5 - 1/2 x) or otherwise rearranged to $2y = 10 - x$ (2y = 10 - x)

PM is equidistant to MQ. The radius from C to M is perpendicular to the chord PQ.

d) Find the coordinates of the point R where the radius intersects the circle.

Right - this is the one. I've tried to substitute the equation of the radius of forms, $2y = 10 - x$ and $y = 5 - 1/2x$ into
$x^2 + y^2 - 12x - 4y + 15$ but I'm not getting the correct answer.

I've tried other approaches like producing a right angle triangle and finding the distance between CM and then from MR, but I don't see how this helps even if I were to use the equation to calculate the distance between C and R, which I've tried with little luck.

I've drawn a diagram and it clearly shows that the coordinates of R are approximately (1.5, 4.2); the actual solution is (6-2sqrt5, 2+sqrt5).

I've looked at other websites and they all suggest that we treat these as simultaneous equations and substitute to find first x and then use the solution of x to find y.

I'd be very grateful if you could please advise if I'm doing anything incorrectly and what I need to do to get the correct solutions please.

Thank you once again for your kind attention.

2. I'll just quickly point out before looking properly that when you said you were using a triangle to find the distance between C and R. Well the distance would just be 5, because that's the radius of your circle.

3. Originally Posted by rideralexmi6
Hello friends,

I'm new on this forum. I'd be very grateful if you could please advise on how to solve this problem: just the approach would be greatly appreciated please - I've been stuck on this for days - I'd be grateful if you could please help me.

I need to find the coordinates on the circumference whereby the radius of the circle bisects a chord and touches that particular point. I've attached two images illustrating the situation.

I've solved the first three parts without any problems, but the last one is really bugging me. The question is actually one of the challenging ones on the exercise:

4. a) Show that the points P(2, -1) and Q(6, 7) lie on the circle whose equation is:

$x^2 + y^2 - 12x - 4y + 15$ <<<<<< this isn't an equation

(again, because I'm new, in case that equation doesn't appear, the equation is x^2 + y^2 - 12x - 4y + 15).

I've done this by subbing P and Q into that equation. PQ therefore form a chord.

b) Find the coordinates of the point M which is the midpoint of the chord PQ.

Midpoint is (4, 3).

c) Find the equation of the radius of the circle that passes through M.

The equation is $y = 5 - 1/2x$ (y = 5 - 1/2 x) or otherwise rearranged to $2y = 10 - x$ (2y = 10 - x)

PM is equidistant to MQ. The radius from C to M is perpendicular to the chord PQ.

d) Find the coordinates of the point R where the radius intersects the circle.

Right - this is the one. I've tried to substitute the equation of the radius of forms, $2y = 10 - x$ and $y = 5 - 1/2x$ into
$x^2 + y^2 - 12x - 4y + 15$ but I'm not getting the correct answer.

I've tried other approaches like producing a right angle triangle and finding the distance between CM and then from MR, but I don't see how this helps even if I were to use the equation to calculate the distance between C and R, which I've tried with little luck.

I've drawn a diagram and it clearly shows that the coordinates of R are approximately (1.5, 4.2); the actual solution is (6-2sqrt5, 2+sqrt5).

I've looked at other websites and they all suggest that we treat these as simultaneous equations and substitute to find first x and then use the solution of x to find y.

I'd be very grateful if you could please advise if I'm doing anything incorrectly and what I need to do to get the correct solutions please.

Thank you once again for your kind attention.
1. I assume that the equation of the circle is

$x^2 + y^2 - 12x - 4y + 15 = 0~\implies~(x-6)^2+(y-2)^2=-15+36+4$

That means the centre is C(6, 2) and the radius is r = 5

2. Calculate the coordinates of the point of intersection between the circle and the line $CM: \dfrac{y-6}{x-2}=\dfrac{3-6}{4-2}=-\dfrac32~\implies~y = -\frac32 x+9$

3.
$x^2 + \left(-\frac32 x+9 \right)^2 - 12x - 4\left(-\frac32 x+9\right) + 15 = 0$

Solve for x. Plug in the x-value into the equation of the line to get the y-coordinate of the point in question.

4. alternatively, using your triangle, i'll call the right angle A
$\frac{1}{\sqrt5}=\frac{RA}{5}$
$\frac{2}{\sqrt5}=\frac{AC}{5}$
(think of SOHCAHTOA for this)
so $RA=\sqrt5$ and $AC=2\sqrt5$
these aren't our coordinates though, we can use the point C to get them though and it gives $(6-2\sqrt5, 2-\sqrt5)$

5. Originally Posted by Rubberduckzilla
I'll just quickly point out before looking properly that when you said you were using a triangle to find the distance between C and R. Well the distance would just be 5, because that's the radius of your circle.
Hello rubberduckzilla

I thought I mentioned that the radius of the circle is indeed 5 in the image? The hypotenuse of that triangle (which is another reconstruction of the problem) is 5 where CM + MR = 5.

I know CM is sqrt5 because:

$sqrt(6 - 4)^2 + (2 - 3)^2 = sqrt5$

Therefore, since I don't know the coordinates of R, I can only assume that the distance MR will be $5 - sqrt5$, where 5 is the distance of the radius subtracting the distance CM.

I don't know how that will actually reconstruct a solution to the problem actually, I only tried to just show if it's a possible approach?!

Thanks again rubberduckzilla

6. Originally Posted by earboth
1. I assume that the equation of the circle is

$x^2 + y^2 - 12x - 4y + 15 = 0~\implies~(x-6)^2+(y-2)^2=-15+36+4$

That means the centre is C(6, 2) and the radius is r = 5

2. Calculate the coordinates of the point of intersection between the circle and the line $CM: \dfrac{y-6}{x-2}=\dfrac{3-6}{4-2}=-\dfrac32~\implies~y = -\frac32 x+9$

3.
$x^2 + \left(-\frac32 x+9 \right)^2 - 12x - 4\left(-\frac32 x+9\right) + 15 = 0$

Solve for x. Plug in the x-value into the equation of the line to get the y-coordinate of the point in question.

Hello earboth

So sorry about the typing error! Yes, the equation was $x^2 + y^2 - 12x - 4y + 15 = 0$

Thanks

7. Originally Posted by Rubberduckzilla
alternatively, using your triangle, i'll call the right angle A
$\frac{1}{\sqrt5}=\frac{RA}{5}$
$\frac{2}{\sqrt5}=\frac{AC}{5}$
(think of SOHCAHTOA for this)
so $RA=\sqrt5$ and $AC=2\sqrt5$
these aren't our coordinates though, we can use the point C to get them though and it gives $(6-2\sqrt5, 2-\sqrt5)$

Actually, the section assumes we don't have any trigonometric knowledge whatsoever so we just have to apply coordinate geometry rules.

I take it you're from UK as well I am - this question actually appears on Core 1 for AQA. I'm not sure if you would happen to have any experience of that maybe?

Thanks

8. Originally Posted by earboth
1. I assume that the equation of the circle is

$x^2 + y^2 - 12x - 4y + 15 = 0~\implies~(x-6)^2+(y-2)^2=-15+36+4$

That means the centre is C(6, 2) and the radius is r = 5

2. Calculate the coordinates of the point of intersection between the circle and the line $CM: \dfrac{y-6}{x-2}=\dfrac{3-6}{4-2}=-\dfrac32~\implies~y = -\frac32 x+9$

3.
$x^2 + \left(-\frac32 x+9 \right)^2 - 12x - 4\left(-\frac32 x+9\right) + 15 = 0$

Solve for x. Plug in the x-value into the equation of the line to get the y-coordinate of the point in question.

Hello earboth

I don't get the solutions matching 6-2sqrt5 or 2 + sqrt5. I get something like 2.4 and 7.8 on my graphics calculator.

Thanks.

9. Originally Posted by Rubberduckzilla
alternatively, using your triangle, i'll call the right angle A
$\frac{1}{\sqrt5}=\frac{RA}{5}$
$\frac{2}{\sqrt5}=\frac{AC}{5}$
(think of SOHCAHTOA for this)
so $RA=\sqrt5$ and $AC=2\sqrt5$
these aren't our coordinates though, we can use the point C to get them though and it gives $(6-2\sqrt5, 2-\sqrt5)$

Hello Rubberduckzilla

Wait !! Whoa!! Could you please explain how you're getting all this because this is the first time I'm actually looking at this and I don't think I've done this stuff yet!

Thanks again mate

10. equation of radius: 2y = 10 - x
equation of circle: $x^2 + y^2 - 12x - 4y + 15 = 0$
solving the two equations you get
$x^2 + \frac{(10 - x)^2}{2^2} - 12x - 2(10-x) + 15 = 0$

$4x^2 + (10 - x)^2 - 40x - 20 = 0$
$4x^2 + 100 - 20x + x^2 - 40x - 20 = 0$
$5x^2 - 60x + 80 = 0$
$x^2 - 12x + 16 = 0$
$x = \frac{12 \pm \sqrt{12^2 - (4)(16)}}{2}$
$x = \frac{12 \pm \sqrt{80}}{2}$
$x = 6 \pm 2\sqrt{5}$
from the graph you see that the for point R,
$x = 6 - 2\sqrt{5}$
substitute it into equation of radius, you have:
$2y = 10 - (6 - 2\sqrt{5})$
$2y = 10 - 6 + 2\sqrt{5}$
$2y = 4 + 2\sqrt{5}$
$y = 2 + \sqrt{5}$
so R = ( $6 - 2\sqrt{5}, 2 + \sqrt{5}$)

11. Originally Posted by ukorov
equation of radius: 2y = 10 - x
equation of circle: $x^2 + y^2 - 12x - 4y + 15 = 0$
solving the two equations you get
$x^2 + \frac{(10 - x)^2}{2^2} - 12x - 2(10-x) + 15 = 0$

$4x^2 + (10 - x)^2 - 40x - 20 = 0$
$4x^2 + 100 - 20x + x^2 - 40x - 20 = 0$
$5x^2 - 60x + 80 = 0$
$x^2 - 12x + 16 = 0$
$x = \frac{12 \pm \sqrt{12^2 - (4)(16)}}{2}$
$x = \frac{12 \pm \sqrt{80}}{2}$
$x = 6 \pm 2\sqrt{5}$
from the graph you see that the for point R,
$x = 6 - 2\sqrt{5}$
substitute it into equation of radius, you have:
$2y = 10 - (6 - 2\sqrt{5})$
$2y = 10 - 6 + 2\sqrt{5}$
$2y = 4 + 2\sqrt{5}$
$y = 2 + \sqrt{5}$
so R = ( $6 - 2\sqrt{5}, 2 + \sqrt{5}$)

Hello ukorov

that made some sense! I see you rearranged 2y = 5 - x into y = 5 - x/2 which I think is a much easier approach and I honestly can't believe I missed this out as well lol! All the other parts are easy, and I can't believe I was stupid enough to miss the rearrangement part out!!!!

Thanks for the simplification and solving the problem - you should have put this icon at the end to show you really know this stuff ... lol

Thanks again!

12. Originally Posted by rideralexmi6
Hello Rubberduckzilla

Wait !! Whoa!! Could you please explain how you're getting all this because this is the first time I'm actually looking at this and I don't think I've done this stuff yet!

Thanks again mate
ok, starting with the notation i've used, on your triangle diagram, you have points R and C I've added a point and called it A where your right angle is on the diagram. If i say a point eg A this refers to that point's angle which in this case would be 90, if i say two points eg RA this refers to the distance between the two points.
If we take $\sin C=\frac{opposite}{hypotenuse}$ (this should be GCSE) you can find one length and then you can use cos for the other. you'll be able to rearange them as $\frac{opposite1}{hypotenuse1}=\frac{opposite2}{hyp otenuse2}$ hope this helps draw a little understanding. =)

13. Originally Posted by Rubberduckzilla
ok, starting with the notation i've used, on your triangle diagram, you have points R and C I've added a point and called it A where your right angle is on the diagram. If i say a point eg A this refers to that point's angle which in this case would be 90, if i say two points eg RA this refers to the distance between the two points.
If we take $\sin C=\frac{opposite}{hypotenuse}$ (this should be GCSE) you can find one length and then you can use cos for the other. you'll be able to rearange them as $\frac{opposite1}{hypotenuse1}=\frac{opposite2}{hyp otenuse2}$ hope this helps draw a little understanding. =)

hello rubberduckzilla