Results 1 to 10 of 10

Math Help - analytical geometry,circles

  1. #1
    Member
    Joined
    Nov 2009
    Posts
    202

    analytical geometry,circles

    please see attachment for qustions:
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,810
    Thanks
    116
    Quote Originally Posted by Pulock2009 View Post

    Find the equation of the circle which touches both axes and passes through
    the point (-2,-1).
    1. The axes touch the circle in the points P(r,0) and Q(0,r). The center of the circle must be situated in the 3rd quadrant and on the line y = x.

    2. The circle has the equation

    (x-a)^2+(y-b)^2=r^2

    According to 1) the coordinates of the center must be equal C(-r , -r):

    3. Plug in the coordinates of P, Q, C and the coordinmates of the given point:

    (-2+r)^2+(-1+r)^2=r^2

    Solve for r.

    4. You'll get 2 circles which satisfy the given conditions.
    Attached Thumbnails Attached Thumbnails analytical geometry,circles-zweikrs_beruehrachsen.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2009
    Posts
    202
    thanks a lot for your answer. one thing i didnot get:how will we know that the centre lies on y=x???and wat about my second question???is it too easy???
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2009
    Posts
    202
    also , will both the circles pass through (-2,-1)???
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,810
    Thanks
    116
    Quote Originally Posted by Pulock2009 View Post
    thanks a lot for your answer. one thing i didnot get:how will we know that the centre lies on y=x???
    The axes are tangents to the circle. The point of intersection of the two tangents and the center of the circle determine the angle bisector between the tangents. The angle bisector of the coordinate axes is the straight line defined by y = x (or y = -x respectively)

    and wat about my second question???is it too easy???
    I've had a lot of trouble to read your attachment. The second question was unreadable for me. Sorry.

    Quote Originally Posted by Pulock2009 View Post
    also , will both the circles pass through (-2,-1)???
    Yes. Have a look at the attachment. The point (-2, -1) is drawn. And you can show that the distance between (-2, -1) and (-5, -5) (or (-2, -1) and (-1, -1) ) is as long as the corresponding radius.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,810
    Thanks
    116
    Quote Originally Posted by Pulock2009 View Post
    please see attachment for qustions:
    Is this the text of the 2nd question?:

    2)Find the area of an equilateral triangle inscribed in the circle x^{2}+y^{2}+2gx+2fy+c=0

    Assuming the points to be A(x_{1},y_{1}),\ B(x_{2},y_{2}),\ C(x_{3},y_{3}) the area of the triangle is:

    ar(ABC)=\begin{array}{ccc}<br />
x_{1} & y_{1} & 1 \\ <br />
x_{2} & y_{2} & 1 \\ <br />
x_{3} & y_{3} & 1<br />
\end{array}
    and there is a similar form of the equation of a circle but i donot know how to combine them for a solution.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2009
    Posts
    202
    yes. from "Assuming....."is the material that i tried out.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,810
    Thanks
    116
    Quote Originally Posted by Pulock2009 View Post
    yes. from "Assuming....."is the material that i tried out.
    1. Transform the equation of the circle such that you can determine the coordinates of the center and the length of the radius:

    x^2+2gx+y^2+2fy=-c~\implies~(x+g)^2+(y+f)^2=f^2+g^2-c

    Thus r^2=f^2+g^2-c

    2. The side of an equilateral triangle which is inscribed in a circle has the length s = r \cdot \sqrt{3}

    The area of such a triangle is A = \frac12 \cdot \underbrace{r \cdot \sqrt{3}}_{\text{length of base}} \cdot \underbrace{\frac32 \cdot r}_{\text{length of height}} = \frac34 \cdot r^2 \cdot \sqrt{3}

    3. Plug in the term for rē from 1. and you're done.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Nov 2009
    Posts
    202
    s=r*sqrt(3)???? how do we get that??? some trigonometry is necessary perhaps????anyways thanks a lot!!!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,810
    Thanks
    116
    Quote Originally Posted by Pulock2009 View Post
    s=r*sqrt(3)???? how do we get that??? some trigonometry is necessary perhaps????anyways thanks a lot!!!
    In this case simple Pythagorean Theorem will do:

    1. See attachment. The small right triangle (coloured in blue) is a 30°-60°-right triangle.

    2. Therefore the blue line segment must have the length \tfrac12 r

    3. The 2nd leg of the right angle has the length:

    \sqrt{r^2-\left(\frac12 r\right)^2} = \frac12 r\cdot \sqrt{3}

    4. This 2nd leg is as long as a half side of the equilateral triangle:

    \frac12 s = \frac12 r\cdot \sqrt{3}~\implies~\boxed{s = r \cdot \sqrt{3}}
    Attached Thumbnails Attached Thumbnails analytical geometry,circles-gleichseitig3eck.png  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Analytical Geometry
    Posted in the Geometry Forum
    Replies: 1
    Last Post: July 15th 2010, 12:08 PM
  2. Analytical Geometry
    Posted in the Geometry Forum
    Replies: 3
    Last Post: May 24th 2009, 10:32 AM
  3. Analytical geometry
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: July 22nd 2007, 09:15 AM
  4. Analytical Geometry help please!
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 10th 2007, 06:57 AM
  5. Analytical Geometry
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: June 20th 2006, 04:00 PM

Search Tags


/mathhelpforum @mathhelpforum