Given that the equation of a circle is x2 + y2-6x-8y = 0, let the longest chord and the shortest chord of the circle passing through the point (3, 5) be AC and BD respectively, then the area of the quadrilateral ABCD is () A. 106B. 206C. 306D. 406

Given that the equation of a circle is x2 + y2-6x-8y = 0, let the longest chord and the shortest chord of the circle passing through the point (3, 5) be AC and BD respectively, then the area of the quadrilateral ABCD is () A. 106B. 206C. 306D. 406

The standard equation of the circle is (x-3) 2 + (y-4) 2 = 52. The longest chord | AC | = 2 × 5 = 10 is obtained from the title. According to the Pythagorean theorem, the shortest chord | BD | = 252 − 12 = 46, and AC ⊥ BD, the area of quadrilateral ABCD s = | 12ac | · | BD | = 12 × 10 × 46 = 206

It is known that the line ι passing through point P (3,8) intersects with circle C: X & sup2; + Y & sup2; - 4y-21 = 0 at two points a and B, and | ab | = 8. The equation of line L is obtained

Circle C: the square of X + the square of y-4y-21 = the center of the circle C (0,2), the radius is 5
Making CD ⊥ AB over C over D
So ad = 4, CD = 3
CD is the distance from C to line ab
Let the slope of the line l be K, then the equation is Y-8 = K (x-3), that is, kx-y + 8-3k = 0
So | 0-2 + 8-3k | / √ (k ^ 2 + 1) = 3
k=3/4
So the equation of line L is 3 / 4x-y + 27 / 4 = 0

It is known that the circle O is circumscribed with the circle C: x2 + Y2 + 6x-8y + 21 = 0 with the origin as the center. (1) find the equation of circle O; (2) find the chord length of the intersection of the straight line x + 2y-3 = 0 and circle o

(1) Let the equation of circle o be x2 + y2 = R2. Circle C: (x + 3) 2 + (y-4) 2 = 4, r = | OC | - 2 = (− 3) 2 + 42 − 2 = 3, so the equation of circle O is x2 + y2 = 9. (2) the distance from O to the straight line x + 2y-3 = 0 is d = 31 + 4 = 355, so the chord length L = 2r2 − D2 = 29 − 95 = 1255