...circle equation? and if (-2,2) is the center of the circle the equation should look like this: (x+2)^2+(Y-2)^2=R^2 And now only R is needed. given 2x-5y+4=0 equation of line perpendicular we can ...
Question about Office Equipment & Supplies
...your answer. x^2 + y^2 = 5 The equation of a circle is (x-h)^2+(y-k)^2=r^2. Since h and k are 0, we have x^2 + y^2=r^2. Since r= 5, r^2=25. Thus, the equation is x^2 + y^2 = 25. Good luck. ...
...a circle How do I enter a circle equation to get the calculator to graph it? Use the conics utility Scroll until you find the equation of a circle and an image, Select it, then enter the
...do I enter a circle equation and get the fx-0750gii calculator to graph it? Use the conics utility Scroll until you find the equation of a circle and an image, Select it, then enter the
Question about Texas Instruments Office Equipment & Supplies
...circle in the regular y= screen, you have to graph it in 2 lines on the y= screen. I assume you've solved for y and gotten a square root equation. Remember a square root can be positive or negative. ...
equation of a circle is aX^2 + aY^2 + bX+cY+d=0 Use the function graphing. But before you can do that you must transform the general equation (above) in such a way that you can write it as Y^2= X^2+EX
circle, on a line parallel to 3x -4y-10=0. The equation of this line is y= 3/4x - 1.75. The y-intercept is -1.75. Now we have two points on the opposite sides of the circle, (1, -1) and (0,-1.75). The
circle equation (x-h)^2+(y-k)^2=r^2
where (h,k) is the centre of the circle and r is the radius of the circle.
circle equation (x-h)^2+(y-k)^2=r^2 where (h,k) is the centre of the circle and r is the radius of the circle. Good luck. Paul
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