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beks73 [17]
2 years ago
14

What is the standard equation for the circle with center (1,-3) that passes through the point (2,2)?

Mathematics
1 answer:
CaHeK987 [17]2 years ago
6 0

The standard equation for the circle is (x - 1 ) ^2 + (y + 3)^2 = 2 ^2

A circle is a closed curve that is drawn from a fixed point called the center in which all points on the curve are the same distance from the center of the center. The equation of a circle with center (h, k) and radius r is given by:

(x-h)^2 + (y-k)^2 = r^2

This is the standard form of the equation. So if we know the coordinates of the center of the circle and also its radius, we can easily find its equation.

Consider any point P(x, y) on the circle. Let "a" be the radius of the circle which is equal to OP.

We know that the distance between the point (x, y) and the origin (0,0) can be found using the distance formula which is equal to -

√[x^2+y^2]= a

Therefore the equation of a circle with center as origin is,

x^2+y^2= a^2

Where "a" is the radius of the circle.

An alternative method

Let's derive another way. Assume that (x,y) is a point on the circle and the center of the circle is at the origin (0,0). Now, if we draw a perpendicular from the point (x,y) to the x-axis, we get a right triangle, where the radius of the circle is the hypotenuse. The base of the triangle is the distance along the x-axis and the height is the distance along the y-axis. Applying the Pythagorean theorem here, we therefore obtain:

x^2+y^2 = radius^2

We need to find the standard equation for the circle with center (1,-3) that passes through the point (2,2)

Standard equation for circle is

(x-h)^2 + (y-k)^2 = r^2................(1)

Here h = 1 , k = -3 and r = 2

put this value of h, k and r in equation (1), we get

(x - 1 ) ^2 + (y + 3)^2 = 2^2

Hence the standard equation of the circle is

(x - 1 ) ^2 + (y + 3)^2 = 2 ^2

<u>Learn more about standard equation of the circle here:</u>

brainly.com/question/28053206<u> </u>

#SPJ4

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