Find the standard form for the equation of a circle ?
Center is the midpoint of the diameter = (-2, -2) Radius = 8, distance from center to either point. (x + 2)^2 + (y + 2) ^ 2 = 64
Find the standard form of the equation of the circle with center (-4,3) and tangent to the line y=1.?
We must first find the distance from point(-4,3) to line y=1 We know that point(-4,1) is on line y=1 and is therefore 2 units away. So our radius is 2. (x-h)^2+(y-k)^2=r^2 is with center(h,k) and radius r. So with center (-4,3) and radius 2 we get (x+4)^2 + (y-3)^2= 4
Find the standard form for the equation of a circle (x-h)^2 + (y-k)^2 = r^2 with a diameter that has endpoints?
find the midpoint of the diameter, that will be the center x = (-5 + 4) / 2 x = -1/2 y = (1 - 8) / 2 y = -7/2 center is (-1/2, -7/2) (x - (-1/2))^2 + (y - (-7/2))^2 = r^2 (x + 1/2)^2 + (y + 7/2)^2 = r^2 find the distance of the diameter and divide by 2 for the radius d = √[(-5 - 4)^2 + (1 - (-8))^2] d = √(81 + 81) d = √162 d = 9√2 r = 9√2/2 r^2 = 162/4 r^2 = 81/2 (x + 1/2)^ + (y + 7/2)^2 = 81/2
Find the equation of a circle in standard form that is tangent to the line x=-3 at (-3,5) and also tangent...?
That sounds more like a command than a question! But I'll play along. Draw a diagram. If the circle is tangent to two different vertical lines, then the distance between the two lines must be the diameter. So the diameter is 9 - (-3) = 12. So the radius is r=6. Now we just need to find the center point of the circle. The distance between the center point and (-3,5) must be the radius, so we need to find (h,k) so that the distance between (h,k) and (-3,5) is 6. Since x= -3 is vertical and (-3,5) is the tangent point, then the center point must be along y= 5. So k=5. This means h is -3+6 = 3. So the center is (3,5). Now you should be able to write the equation.
How to find the standard form of the equation of the specified circle?
do you know the pythagoreum theorum? Lets create this from the beginning like the guy who invented circles by thinking about it. Ready A^2+B^2=C^2 Right triangle and hypotenuse. Well what if our hypotenuse was always nine the radius of your circle? And we had a million little triangles of all types with a radius/hypotenuse of 9. Each triangle with hypotenuse/radius of nine would have every combination of legs A and B that gives you hypotenuse of 9. What if leg a of your triangle was always your x value on for each point that makes up this circle. What if leg B of your triangle was always your y value on for each point that makes up this circle. You would find (substituting x and y in for a and b) that x^2+y^2=r^2 if you graph that x^2 + y^2 always equal 81 Which will draw a perfect circle and help you learn the equation for a circle....
Find the standard form of the equation of a circle having the following properties?
Center at the point (4,-3) Tangent to the y axis. Type the standard form of the equation of the circle (h, k) = (4, –3) Since it is tangent to the y-axis, radius = h = 4 Equation of circle: (x – h)² + (y – k)² = r² (x – 4)² + (y + 3)² = 4²
How do you find the Standard Equation of a Circle whose center is at (-3,2) and is tangent to a line whose equation is 2x+3y=1?
How do you find the Standard Equation of a Circle whose center is at (-3,2) and is tangent to a line whose equation is 2x+3y=1?A2AAs the circle has a center (-3, 2), its equation will be (x - -3)^2 + (y - 2)^2 = r^2. So we just need to find r. If r is too small, it will not intersect the line. If it is too big, it will intersect the line twice, so we need to find the r that makes it “just right” (a la Goldilocks). Think about if you were at a point, (like (-3, 2)), and you were trying to get to a line (like 2x + 3y = 1). What would be the quickest way? Well, the slowest way would be to walk parallel to the line — you’d never get there. The quickest way is to walk towards the line along a line with is perpendicular to the line itself. So let’s find the slope of 2x + 3y = 1.2x + 3y = 13y = -2x + 1y = (-2/3)x + 1/3.The line which is perpendicular to this line has a slope which is the additive inverse of the multiplicative inverse of -2/3. The multiplicative inverse of -2/3 is -3/2, and the additive inverse of -3/2 is 3/2.Aside: You can say “negative” vs. additive inverse, and “reciprocal” instead of multiplicative inverse, but the difference is just a single word, and it stresses both that you are dealing with the idea of inverses, and which operations you are dealing with. It is a great habit to get into to use these two phrases. Also, “opposite” is just a really terrible word to use for mathematics.So we are going to find the equation of the line that goes through (-3, 2), and has the slope 3/2. Then we’ll see where it intersects y = (3/2)x +13/2, and then find the length of that line segment between (-3, 2) and this intersection.y - 2 = 3/2(x - -3) (point slope form)y = (3/2)x +9/2 + 4/2y = (3/2)x +13/2Setting the y values of these two lines equal to one another yields:(3/2)x +13/2 = (-2/3)x + 1/3(3/2 + 2/3)x = 1/3 - 13/2(9/6 + 4/6)x = 2/6 - 39/6Multiply both sides by 6 and simplify13x = -37x = -37/13.So when x = -37/13, the y values of BOTH lines are equal, so I can put this value of x into either line to find the y value of where these lines intersect.y =(-2/3)(-37/13) + 1/3y = 74/39 +13/39y = 87/39 = 29/13.So, finally, the distance between (-3, 2) and (-37/13, 29/13) is radical( (-2/13)^2 + (3/13)^2) = radical(13/169) = radical(1/13),So, the equation of this circle is:(x + 3)^2 + (y - 2)^2 = 1/13
Find the equation in standard form of a circle that satisfies the given condition.?
radius = (3--5)=8 therefore (x--5)^2 + (y-1)^2 = 8^2 therefore (x+5)^2 + (y-1)^2 = 64