) RRRy dA; R is the region in the first quadrant enclosed between the circle x2 + y2 = 25and the line x + y = 5

evaluate double integral y dydx where r is the region in first quadrant enclosed between the circle x^2+y^2=25 and straight line x+y=25

Find the area of the region.interior of r = 5 + 3 sin(𝜃) (below the polar axis

Which circle does the point (-1, 1) lie on?(x-5)2 + (y+2)2 = 25(x-2)2 + (y+6)2 = 25(x-2)2 + (y-5)2 = 25(x-2)2 + (y-2)2 = 25

find the coordinates of the points of intersaction of the straight line of the staight line5x-4y+25=0 and the circle 2x^2+2y^2+7x-6Y-15=0

To determine the equation of the circle graphed, we need to identify the center and the radius of the circle. 1. **Identify the center of the circle:** From the graph, the center of the circle is at the origin \((0, 0)\). 2. **Determine the radius of the circle:** The radius is the distance from the center to any point on the circle. From the graph, the circle intersects the x-axis at \(x = 5\) and \(x = -5\), and the y-axis at \(y = 5\) and \(y = -5\). Therefore, the radius \(r\) is 5 units. The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is: \[ (x - h)^2 + (y - k)^2 = r^2\] Since the center is \((0, 0)\) and the radius is 5, the equation of the circle is: \[ (x - 0)^2 + (y - 0)^2 = 5^2\] Simplifying, we get: \[ x^2 + y^2 = 25\] So, the equation of the circle is: \[ x^2 + y^2 = 25\]