Maths The Circle

2Fast2C

Could someone help me with these 2 qs.

Q8
A circle of radius lenght 5 units has its centre in the first quadrant.
it touches the x axis and makes an intercept of length 6 units on the y axis.
Find the co ordinates of centre C

Q9.
A circle has centre (2,3) and radius 4 units in lenght.
Show that the distance between he ppojnts where the circle interescts the y axis is 4sqareroot3.
find the lenght of the intercept the circle cuts off the x axis

skippy1977

Make sure to always draw a decent sketch in the circle question. Will help alot. The first question is solved using pythagoras.

skippy1977

The second question is quite similar and uses Pythagoras too. Again a good sketch is the key.

A97

For the second one, where it cuts the y-axis, x is 0. So you have the point (0,y), the centre (2,3) and radius 4. You can solve for y using the equation of the circle. You should get 2 y values. Use the distance formula to verify that the distance between the two is 4√3. Do the same for the x-axis. You're looking for an answer of 2√7.:)

Interesting alternative method up there skippy.

2Fast2C

A97 wrote: »

For the second one, where it cuts the y-axis, x is 0. So you have the point (0,y), the centre (2,3) and radius 4. You can solve for y using the equation of the circle. You should get 2 y values. Use the distance formula to verify that the distance between the two is 4√3. Do the same for the x-axis. You're looking for an answer of 2√7.:)

Interesting alternative method up there skippy.

I am not sure I understand what you mean.
(0,y)
(x-0)^2+(y-y)^2=4^2
x^2=16
x=4

(2,3)
(x-2)^2+(y-3)^2=4^2
x^2+y^2-4x-6y-3

What am I supposed to do with them

Peregrine

2Fast2C wrote: »

I am not sure I understand what you mean.
(0,y)
(x-0)^2+(y-y)^2=4^2
x^2=16
x=4

(2,3)
(x-2)^2+(y-3)^2=4^2
x^2+y^2-4x-6y-3

What am I supposed to do with them

The last line you have there is the equation of the circle.

To find out where it cuts the y axis, let x = 0 and solve for y. You will get two answers (two places where it cuts the y axis). Get the distance between the two co-ordinates using the distance formula.

For the x axis, let y = 0, solve for x and get the distance again.

skippy1977

The method the lads are outlining is perfectly valid. Unfortunately the points where the circle cut the y-axis are not integers. Solving y^2-6y-3 (using - b formula) you'll get the points (0, -0.4641..) and (0, 6.4641...)....also no need to use a distance formula when both points are on the axis. -0.4641 to 6.4641 is 6.9282... units....approximately 4root3.
Much better in this case (though not all cases) to stick to the graphical approach using Pythagoras outlined above.

2Fast2C

I got it lads, thank you for the help.

Peregrine

skippy1977 wrote: »

The method the lads are outlining is perfectly valid. Unfortunately the points where the circle cut the y-axis are not integers. Solving y^2-6y-3 (using - b formula) you'll get the points (0, -0.4641..) and (0, 6.4641...)....also no need to use a distance formula when both points are on the axis. -0.4641 to 6.4641 is 6.9282... units....approximately 4root3.
Much better in this case (though not all cases) to stick to the graphical approach using Pythagoras outlined above.

Oops, yeah. No need for the distance formula at all.

I prefer the Pythagoras way too. I guess solving for x and y at the axes is more methodical if you know what I mean? Some students are inclined to think like that. I usually always prefer the graphical approach myself.