Find the center and radius of the circle \(x^2+y^2-4x-8y-45=0\)
Solution and Explanation
Given that
The equation of the circle \(x^2+y^2-4x-8y-45=0\)
Then we know that The standard equation of the circle is \((x - h)^2 + (y -k)^2 = r^2\)
\(x^2+y^2-4x-8y-45=0\)
\(⇒x^2+2^2-4x+y^2+4^2-8y-20-45=0\)
\(⇒x^2+2^2-4x+y^2+4^2-8y-65=0\)
\(⇒(x-2)^2+(y-4)^2=(√65)^2\)
Now comparing the above equation with the standard equation we get
The circle of the equation is \((2,4)\) and the radius is \(√65\)
So by comparing the standard equation with the given equation, we get
\(h=-5 , k=3\) and \(r=6\)
Hence the center of the circle is \((-5,3)\), and the radius is \(6\).
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Concepts Used:
Circle
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
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