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Please help me with the following problem:
Write down the equation of a circle radius 2 and center (1,1). Find the points where this circle cuts the axes. Find the equation of the tangent to this circle at the point where it cuts the positive y-axis and find the point where this tangent cuts the x-axis.
Thx in advance :)
some info you can use to help solve this ...Quote:
Originally Posted by Boz
circle with radiuscentered at
is
the circle intersects the y-axis when
the slope of a tangent line to a circle is perpendicular to the slope of the radius at the point of tangency.
Hi Boz,
You have not replied to Skeeters post.Did you work it out? If not here are suggestions.
1 plot the circle
2Label as follows
A negative x intersection of circle
B positive x " " "
C negative y " " "
D positive y " " "
3 note the slope diagrams which can be formed by drawing radii to each intersection with appropriate horizontal or vertical lines
4 What are the properties of a 2-1 right triangle?
5 carry on
bjh
I haven't worked it out yet.But it's part of my homework for Monday, so I'd be grateful if you write me the full solution.Quote:
Originally Posted by bjhopper
Im gonna learn how to do these kind of problems,because my exam is in one month time.However,now Im revising calculus.
You will learn much better by doing it yourself rather than having someone else give you the "full solution". Skeeter told you that the equation of a circle with center at (h, k) and radius r is.
You are told that the center of this circle is at (1, 1) and its radius is 2. What is its equation?
That circle cuts the x-axis where y= 0 and the y-axis where x= 0. Put y= 0 in the equation and solve for x. Then put x= 0 and solve for y. (Do not multiply out the squares to solve the equations. You should be able to write them asand
and take the square root of both sides.)
A tangent line is always perpendicular to the radius of the circle. What is the slope of the line from (1, 1) to the point where the circle cuts the positive x-axis? The slope of the tangent line is the negative of the reciprocal of that (-1/m). What is the equation of the tangent line? Set y= 0 in the equation of that tangent line and solve for x.
Hi Boz,
It would be wrong for me to do your homework.Show your work and then ask for help but never full solutions.
bjh
(x-1)^2 + (y-1)^2 =2^2Quote:
Originally Posted by bjhopper
The circuit intersects y-axis when x=0.Then (0-1)^2+(y-1)^2=4
y=1+sqrt 3 = 2.73
The same goes for the x-axis. x= 2.73
I dont know how to continue.
Hi Boz,
+x and +y correct. I assume you can also provide -x and -y values.What is the slope of OB and then the slope of the tangent at B. Write the equation for the tangent using the point slope formula.
bjh
Better shut down this forum (Hi)Quote:
Originally Posted by bjhopper