Check the position of the Point (5,6) with respect to its circle 2x²+2y²+12x-8y+1=0

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Check the position of the Point  (5,6) with respect to its circle  2x²+2y²+12x-8y+1=0

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Convert to Standard Form: Convert the given circle equation into the standard form. The standard form of a circle's equation is (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is the radius. Given equation: 2x² + 2y² + 12x - 8y + 1 = 0 Divide the entire equation by 2 to simplify: x² + y² + 6x - 4y + 1/2 = 0Complete the Square: Complete the square for both x and y. For x: Take half of the coefficient of x, square it, and add it to both sides: (x + 3)² - 9 For y: Take half of the coefficient of y, square it, and add it to both sides: (y - 2)² - 4 Now add 9 and 4 to both sides to balance the equation: (x + 3)² + (y - 2)² = 9 + 4 - 1/2Simplify Equation: Simplify the right side of the equation. 9 + 4 - 1/2 = 13 - 1/2 = 12.5 So the equation of the circle in standard form is: (x + 3)² + (y - 2)² = 12.5Determine Center and Radius: Determine the center (h, k) and the radius r of the circle. From the standard form, we have h = -3, k = 2, and r² = 12.5. 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. Substitute x = 5 and y = 6 into the standard form: (5 + 3)² + (6 - 2)² = 12.5 Calculate: (8)² + (4)² = 12.5 64 + 16 = 12.5 80 ≠ 12.5Compare Results: Compare the result with r² to determine the position of the point. Since 80 > 12.5, the point (5,6) lies outside the circle.

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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Check the position of the Point (5,6)(5,6)(5,6) with respect to its circle 2x2+2y2+12x−8y+1=02x^2+2y^2+12x-8y+1=02x2+2y2+12x−8y+1=0

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