2016年4月8日 · The polar equation of a full circle, referred to its center as pole, is r = a. The range for #theta# for the full circle is #pi#. For half circle, the range for #theta# is restricted to #pi#. So, the answer is r = a and #alpha < theta < alpha + pi#, where a and #alpha# are constants for the chosen half circle.

A circle C has equation #x^2+y^2-6x+8y-75=0#, and a second circle has a centre at #(15,12)# and radius 10. What are the coordinates of the point where they touch? If a circle has center (0,0) and a point on the circle (-2,-4) write the equation of the circle.?

2018年7月13日 · The graph is a circle so all the points are enclosed in it. The domain is the values for x so you subtract the radius from the centre coordinate and you add the radius to it. The range is the values for y so you do the same to the y coordinate. If you use #(x+2)^2+(y-4)^2=25# The centre is (-2,4) radius is 5. Domain #=> -2-5=-7# and # -2+5=3 ...

2014年8月23日 · We'll start with the parametric equations for a circle: y = rsin t x = rcos t where t is the parameter and r is the radius. If you know that the implicit equation for a circle in Cartesian coordinates is x^2 + y^2 = r^2 then with a little substitution you can prove that the parametric equations above are exactly the same thing. We will take the equation for x, and solve for t in …

2014年9月14日 · The general form of a circle looks like ... x^2+y^2+Ax+By+C=0 In the standard form to the equation for a circle look like ... (x-h)^2+(y-k)^2=r^2 sqrt(r^2)=r,radius Convert the general form to standard form by using the completing the square process. You will then have the r^2 value. The square root of r^2 is the radius of the circle.

2018年5月3日 · Given: circle equation #x^2 + y^2 + 4x - 8y + 13 = 0# The standard form of circle equation is: #(x-h)^2 + (y-k)^2 = r^2# where center # = (h, k)#, #" "r = # radius. Use completing of the square to find the equation of the circle: First combine the #x# terms and the #y# terms and put the constant(s) on the right side of the equation:

2015年11月17日 · The centre of the original circle is #(0, 0)#. The centre for our shifted circle is #(-5, -2)#. Just replace #x# with #x+5# and #y# with #y+2# in the original equation to get: #(x+5)^2+(y+2)^2 = 19# In fact if #f(x, y) = 0# is the equation of any curve, then #f(x+5, y+2) = 0# is the equation of the same curve shifted left #5# units and down #2 ...

2016年8月2日 · The Centre is (-2,4) and radius=8. Let us rewrite the given eqn. of circle as, x^2+4x+y^2-8y=44, and, complete the squares on L.H.S.. :. x^2+4x+4+y^2-8y+16=44+4+16=64. :. (x+2)^2+(y-4)^2=8^2. We compare this eqn. with : (x-h)^2+(y-k)^2=r^2, representing the circle with centre (h,k) and radius r. Therefore, the Centre is (-2,4) and radius=8.

2018年8月7日 · The center is =(3,3) and the radius is =2 The general equation of a circle, center C=(a,b) and radius =r is (x-a)^2+(y-b)^2=r^2 Here, we have x^2+y^2-6x-6y+14=0 Rearrange the equation and complete the square x^2-6x+y^2-6y=-14 x^2-6x+9+y^2-6y+9=-14+9+9 (x-3)^2+(y-3)^2=4=2^2 The center is =(3,3) and the radius is =2 See the graph below. graph{(x^2+y^2-6x …

2017年1月12日 · This is the equation of a circle, center (-4,3) and radius =5 We need (a+b)^2=a^2+2ab+b^2 (a-b)^2=a^2-2ab ...