Check the position of the Point $(5,6)$ with respect to its circle $2x^2+2y^2+12x-8y+1=0$

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Algebra 2

Find the radius or diameter of a circle

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Q. Check the position of the Point

(5,6)

with respect to its circle

2x^2+2y^2+12x-8y+1=0

Convert to Standard Form: Convert the given circle equation into the standard form. $\newline$ The standard form of a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius. $\newline$ Given equation: $2x^2 + 2y^2 + 12x - 8y + 1 = 0$ $\newline$ Divide the entire equation by $2$ to simplify: $x^2 + y^2 + 6x - 4y + \frac{1}{2} = 0$
Complete the Square: Complete the square for both x and y. $\newline$ For x: Take half of the coefficient of x, square it, and add it to both sides: $(x + 3)^2 - 9$ $\newline$ For y: Take half of the coefficient of y, square it, and add it to both sides: $(y - 2)^2 - 4$ $\newline$ Now add $9$ and $4$ to both sides to balance the equation: $(x + 3)^2 + (y - 2)^2 = 9 + 4 - \frac{1}{2}$
Simplify Equation: Simplify the right side of the equation. $\newline$ $9 + 4 - \frac{1}{2} = 13 - \frac{1}{2} = 12.5$ $\newline$ So the equation of the circle in standard form is: $(x + 3)^2 + (y - 2)^2 = 12.5$
Determine Center and Radius: Determine the center $(h, k)$ and the radius $r$ of the circle. $\newline$ From the standard form, we have $h = -3$ , $k = 2$ , and $r^2 = 12.5$ . $\newline$ Therefore, the center of the circle is $(-3, 2)$ and the radius is $\sqrt{12.5}$ .
Plug in Coordinates: Plug the coordinates of the point $(5,6)$ into the circle's equation to determine its position. $\newline$ Substitute $x = 5$ and $y = 6$ into the standard form: $(5 + 3)^2 + (6 - 2)^2 = 12.5$ $\newline$ Calculate: $(8)^2 + (4)^2 = 12.5$ $\newline$ $64 + 16 = 12.5$ $\newline$ $80 \neq 12.5$
Compare Results: Compare the result with $r^2$ to determine the position of the point. $\newline$ Since $80 > 12.5$ , the point $(5,6)$ lies outside the circle.