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    Home»Others»What Is The Radius Of A Circle Whose Equation Is X2+y2+8x−6y+21=0?
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    What Is The Radius Of A Circle Whose Equation Is X2+y2+8x−6y+21=0?

    GiannaBy GiannaMay 3, 2023Updated:May 3, 2023No Comments4 Mins Read
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    What Is The Radius Of A Circle Whose Equation Is X2+y2+8x−6y+21=0?
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    A circle is a two-dimensional shape with a curved line around its circumference. The radius of a circle is a measure of its size, and it is the distance from the center of the circle to any point on its circumference. The equation of a circle is a mathematical expression that describes the location and size of a circle. In this article, we will be looking at how to calculate the radius of a circle whose equation is x2+y2+8x−6y+21=0.

    What is the Radius of a Circle?

    The radius of a circle is the distance from the center of the circle to any point on its circumference. It can be calculated using the formula r = √(x2 + y2), where x and y are the coordinates of the center of the circle. The radius of a circle is also equal to half of its diameter.

    Solving for the Radius of X2+y2+8x−6y+21=0

    To solve for the radius of a circle whose equation is x2+y2+8x−6y+21=0, we first need to find the equation’s center. To do this, we can set the equation equal to zero and then solve for x and y.

    We can start by subtracting 21 from both sides of the equation to get x2+y2+8x−6y = -21. We can then divide both sides of the equation by 2 to get x2+y2+4x−3y = -10.5.

    Next, we can subtract x2+y2 from both sides of the equation to get 4x−3y = -10.5-x2-y2. We can then divide both sides of the equation by 4 to get x – (3/4)y = -(10.5/4) – (x2/4) – (y2/4).

    Finally, we can subtract (3/4)y from both sides of the equation to get x = -(10.5/4) – (x2/4) – (y2/4) – (3/4)y. We can then solve for x to get x = -(10.5/4) – (3/4)y.

    We can now plug this value for x into the original equation and solve for y. We can start by subtracting 8x from both sides of the equation to get

    When dealing with a circle, the equation that must be solved is of the form (x–a)² + (y–b)² = r² where a and b represent the coordinates of the center of the circle and r represents its radius.

    Given the equation x² + y² + 8x–6y + 21 = 0, one can determine the radius of the circle through several steps.

    First, the equation can be rearranged to look like the generic circle equation. By adding 6y² to both sides and subtracting 21 from both sides, you will have x² + (y² + 8x + 6y) + 21 = 6y².

    Then, using the algebraic property that when two terms are being added or subtracted, the terms can be “flipped” and the sign changed without changing the equation, you can rearrange and combine the terms to get (x + 4)² + (y + 3)² = r².

    From this, you can determine that the center of the circle is located at (–4, –3). The next step is to calculate the radius.

    The equation now simplifies to r² = (x + 4)² + (y + 3)², which after some basic manipulation, yields the algebraic equation r = √((x + 4)² + (y + 3)²). Using this equation, you can find the radius of the circle for a given x and y value. For example, if x = 3 and y = 4, the radius would equal r = √((3 + 4)² + (4 + 3)²) = √46.

    In conclusion, the radius of a circle whose equation is x² + y² + 8x–6y + 21 = 0 can be determined by rearranging the equation to take the generic circle equation form, finding the center of the circle, and then using the equation r = √((x + 4)² + (y + 3)²) to calculate the radius for a given x and y value.

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