Write an equation in standard form of the circle with the given properties. Center at the origin;r=squareroot?
======= The standard form of an equation for a circle with center (h,k) and radius r is (x-h)^2 + (y-k)^2 = r^2 ============ 1: Circle with the center at the origin, (0,0) and radius √5: x^2 + y^2 = 5. ========= 2: Center at (0,8) and r = 8. x^2 + (y-8)^2 = 64 ======== 3. Center at (−15, 0); r = sqaure root 14 (x+15)^2 + y^2 = 14
Write the equation of the circle in standard form x^2+y^2-6x+4y-3+0?
x^2+y^2-6x+4y-3+0..............Let's make the + 0 an = 0 x^2 + y^2 - 6x + 4y - 3=0 (x² - 6x) + (y² + 4y) = + 3 (x² - 6x + ?) + (y² + 4y + ??) = 3 + ? + ?? ? = [ (1/2)(- 6) ]² = ( - 3 )² = 9 = ? ?? = [ (1/2)(4) ]² = ( 2 )² = 4 = ?? (x² - 6x + 9) + (y² + 4y + 4) = 3 + 9 + 4 (x - 3)² + (y + 2)² = 16.............ANSWER Center........( +3 , -2 ) Radius.....√16.......4 Let me know if you have questions on this problem or any other math problems. I am an excellent math teacher/tutor specializing in making students faster.....smarter......and better looking like myself. Okay, just faster and smarter....you don't want to look like me. I TUTOR FREE ONLINE VIA SKYPE. Keeps me busy since moving from Florida to France. I make C, D , F students into A, B , C students. I make great students even greater. If interested just let me know how to get in touch. If not that is fine also. Mathcerely, Robert Jones "teacher of fine students"
Write the standard form of the equation of the circle with the given characteristics. endpoints of a diameter:?
the center will be the midpoint: (3, 4) and r² will be 3² + 4² = 25 so (x - 3)² + (y - 4)² = 25 ♣♦
Write the equation of the circle in standard form?
2x^2 + 2y^2 - 2x - 2y - 3 = 0 2x^2 - 2x + 2y^2 - 2y = 3 (regroup) 2(x^2 - x + ___) + 2(y^2 - y + ___) = 3 + 2(___) + 2(___) 2(x^2 - x + 1/4) + 2(y^2 - y + 1/4) = 3 + 2(1/4) + 2(1/4) 2(x - 1/2)^2 + 2(y - 1/2)^2 = 4 (x - 1/2)^2 + (y - 1/2)^2 = 2 center = (1/2, 1/2), r = sqrt(2)
Write the equation in standard form of the circle with the given properties?
x² + y² = 2 because... the standard from of the equation of a circle is.. (x - h)² + (y - k)² = r² where point(h,k) is the center and r is the radius. The origin is at point(0,0). This problem tells us that r = √2 (x - 0)² + (y - 0)² = (√2)² (x)² + (y)² = 2 x² + y² = 2
Write the equation, in standard form, of the circle with radius 4 and center (-1, 0)?
The equation of a circle is as follows:[math](x-h)^2+(y-k)^2=r^2[/math]Here, [math](h,k)[/math] is the centre of the circle and [math]r[/math] is the radius.On comparing with [math](-1,0)[/math], we get,[math]h=-1[/math] and [math]k=0[/math] along with [math]r=4[/math].Substituting these values in the equation, we get,[math](x-(-1))^2+(y-0)^2=(4)^2[/math][math](x+1)^2+y^2=16[/math]This is the equation of the circle with centre [math](-1,0)[/math] and radius [math]4[/math] which can further be expressed as[math]x^2+2x+1+y^2=16[/math][math]x^2+y^2+2x-15=0[/math]Hope this helps! You can also get such questions solved and get the detailed solution within seconds using the Scholr app by just uploading a picture of the question and also get to be a part of an ever-growing community of students.
Write the standard form of the equation for the circle that passes through the points given coordinates?
Write the standard form of the equation for the circle that passes through the points (–9, –16), (–9, 32), and (22, 15). Then identify the center and radius. Choices: A. (x – 3)^2 + (y + 9)^2 = 25 (–2, 8), 25 B. (x + 2)^2 + (y – 8)^2 = 625 (–2, 8), 25 C (x + 2)^2 + (y – 8)^2 = 625 (–3, 9), 5 D. (x – 3)^2 + (y + 9)^2 = 25 (–3, 9), 5 Please show work Thanks :) Will do best answer! Sal
How do you write the equation of a circle in standard form?
2x^2 + 2y^2 + 4x - 12y +11 = 0 divide by 2 x^2 + y^2 + 2x - 6y +11/2 = 0 x^2 + 2x + y^2 - 6y +11/2 = 0 x^2 + 2x +1 + y^2 - 6y +11/2 = 0 + 1 (x+1)^2 + y^2 - 6y + 9 +11/2 = 0 + 1 + 9 (x+1)^2 + (y-3)^2 + 11/2 = 0 + 1 + 9 (x+1)^2 + (y-3)^2 = 0 + 1 + 9 - 11/2 (x+1)^2 + (y-3)^2 = 9/2 Hope this helps!
What are the equations of at least 2 circles in standard form?
A circle with equation [math]{(x-a)}^{2} + {(y-b)}^{2} = {r}^{2}[/math] has a centre of [math](a,b)[/math] and a radius of [math]r[/math].If the circle's centre is on the x or y axis, either [math]a = 0[/math] or [math]b=0[/math].If the circle's centre is in quadrant 2 or 4, either [math]a > 0[/math] and [math]b < 0[/math] or [math]a < 0[/math] and [math]b > 0[/math].Hope this helps!
What is the equation in standard form for the circle with radius 11 centered at the origin?
Let's start with the basics.What is a circle?A circle is a set of all points in a plane which are at a fixed distance , called radius, from a fixed point, called the centre.So, let's begin with the answer.First of all, what we have been asked to do is to find an equation that represents all points at a fixed distance, say r , from a fixed point, say (a,b).Suppose the point on the locus that we require is (x,y).We know that every point (x,y) is a fixed distance, say r , from a fixed point, say (a,b).Hence, now we apply the distance formula.D[math]=√{((x-a)^2+(y-b)^2)} [/math]This I wrote owing to the Pythagoras theorem.Now; here we have D=radius r. So let's substitute and Square both sides.[math]r^2=(x-a)^2+(y-b)^2[/math]This is the equation of a circle centred at (a,b) and having radius r. Now substitute values; r=11; a=b=0.To get[math] x^2+y^2=121.[/math]This is the required equation.In this way we can find equation of any curve.For practice try solving the following.A line with slope 2 intercepts the co-ordinate axes at A and B. Find the locus of all points (x,y) that are midpoints of segment AB.Circles with radius 2 always pass through the point (3,4). Find the locus of centres of these circles.A tangent to a special parabola is of the form [math]yt=x+at^2 [/math]where t can take any real value and a is fixed constant ( Say 1). Find the locus of the midpoint of the portion intercepted by tangents between the axes.Let me know if you understood and if you could solve the problems.For help, use DM.THANK YOU.Edit. Try learning the approach : not the formula