2<\/sup> terms and these terms have the same coefficient (usually 1).<\/p>\nWhen the equation of a circle appears in “general form”, it is often beneficial to convert the equation to “center-radius” form to easily read the center coordinates and the radius for graphing. \n <\/p>\n
Examples:<\/strong><\/p>\n1. Convert x2<\/sup>+ y2<\/sup>-4x-6y+8= 0\u00a0 into center-radius form.<\/strong><\/p>\nThis conversion requires use of the technique of completing the square.<\/p>\n
We will be creating two perfect square trinomials within the equation.<\/p>\n
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\nStart by grouping the x related terms together and the y related terms together. Move any numerical constants (plain numbers) to the other side.<\/li>\n Get ready to insert the needed values for creating the perfect square trinomials. Remember to balance both sides of the equation.<\/li>\n Find each missing value by taking half of the “middle term” and squaring. This value will always be positive as a result of the squaring process.<\/li>\n Rewrite in factored form. \nYou can now read that the center of the circle is at (2, 3) and the radius is \/5<\/li>\n<\/ul>\n2. How do the coordinates of the center of a circle relate to C and D when the equation of the circle is in the general form x2<\/sup>+ y2<\/sup>+Cx+Dy+E = 0.<\/strong><\/p>\nLet’s make some observations. Re-examine our previous equations in general form and center-radius form. Do you see a relationship between the center coordinates and C and D?<\/p>\n
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3. Write the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).<\/strong><\/p>\n <\/p>\n
4. Write the equation for the circle shown below if it is shifted 3 units to the right and 4 units up.<\/strong><\/p>\nA shift of 3 units to the right and 4 units up places the center at the point (3, 4). The radius of the circle can be seen from the graph to be 5. \nEquation:(x-3)2 <\/sup>+ (y-4)2\u00a0 <\/sup>= 25<\/p>\n5. Convert 2x2<\/sup>+ 2y2<\/sup>+6x-8y+12= 0 into center-radius form.<\/strong><\/p>\nWhoa!!! This equation looks different. Are we sure this is a circle??? \nIn this equation, both the x and y terms appear in squared form and their coefficients (the numbers in front of them) are the same. Yes, we have a circle here! We will, however, have to deal with the coefficients of 2 before we can complete the square.<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
Equation of Circles Let’s review what we already know about circles. Definition: A circle is a locus (set) of points in a plane equidistant from a fixed point. Now, if we “multiply out” the above example (x-2)2 + (y+5)2\u00a0 = 9 we will get: (x-2)2 + (y+5)2\u00a0 = 9 (x2-4x+4) +( y2+10y+25) = 9 x2-4x+4+ … Read more<\/a><\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[3351],"yoast_head":"\nEquation of Circles - CBSE Library<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n