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...**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 ...

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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. ...

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...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**. Good luck. Paul

**circle** **equation** I have a circumference **circle** that is 22 1/2". How do I find the diameter? The circumference (C) of a **circle** and its diameter (D) are related by C=PI*D (Pi is the famous ...

**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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