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Write the standard equation of a circle that passes through the point [tex]\((2,2)\)[/tex] with its center at [tex]\((1,-3)\)[/tex].

A. [tex]\((x+1)^2+(y-3)^2=26\)[/tex]
B. [tex]\((x-1)^2+(y+3)^2=26\)[/tex]
C. [tex]\((x-1)^2+(y-3)^2=26\)[/tex]
D. [tex]\((x-3)^2+(y+1)^2=26\)[/tex]


Sagot :

To determine the standard equation of the circle passing through the point [tex]\((2, 2)\)[/tex] with its center at [tex]\((1, -3)\)[/tex], we will follow these steps:

1. Identify the components of the standard equation of a circle:
The general equation of a circle with center [tex]\((h, k)\)[/tex] and radius [tex]\(r\)[/tex] is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
In this case, the center of the circle is given as [tex]\((1, -3)\)[/tex], so [tex]\(h = 1\)[/tex] and [tex]\(k = -3\)[/tex].

2. Calculate the radius:
To find the radius [tex]\(r\)[/tex], we use the distance formula between the center [tex]\((1, -3)\)[/tex] and the point [tex]\((2, 2)\)[/tex] that lies on the circle:
[tex]\[ r^2 = (x_1 - h)^2 + (y_1 - k)^2 \][/tex]
Substituting the given points [tex]\((x_1, y_1) = (2, 2)\)[/tex], [tex]\(h = 1\)[/tex], and [tex]\(k = -3\)[/tex]:
[tex]\[ r^2 = (2 - 1)^2 + (2 - (-3))^2 \][/tex]
This simplifies to:
[tex]\[ r^2 = 1^2 + 5^2 = 1 + 25 = 26 \][/tex]

3. Construct the equation using the calculated radius:
Now that we know [tex]\(h = 1\)[/tex], [tex]\(k = -3\)[/tex], and [tex]\(r^2 = 26\)[/tex], we can write the standard equation of the circle:
[tex]\[ (x - 1)^2 + (y + 3)^2 = 26 \][/tex]

Thus, the standard equation of the circle that passes through the point [tex]\((2, 2)\)[/tex] with its center at [tex]\((1, -3)\)[/tex] is:
[tex]\[ (x - 1)^2 + (y + 3)^2 = 26 \][/tex]

Therefore, the correct answer is:

[tex]\((x - 1)^2 + (y + 3)^2 = 26\)[/tex]