The known equation x ^ 2 + y ^ 2-2 (T + 3) + 2 (1-4t ^ 2) y + 16t ^ 4 + 9 = 0 represents a circle and finds the circle with the largest area

The known equation x ^ 2 + y ^ 2-2 (T + 3) + 2 (1-4t ^ 2) y + 16t ^ 4 + 9 = 0 represents a circle and finds the circle with the largest area

X ^ 2 + y ^ 2-2 (T + 3) x + 2 (1-4t ^ 2) y + 16t ^ 4 + 9 = 0, i.e. [x - (T + 3)] ^ 2 + [y + (1-4t ^ 2)] ^ 2 = - 7T ^ 2 + 6T + 1 〈 R ^ 2 = - 7T ^ 2 + 6T + 1 = - 7 (T-3 / 7) ^ 2 + 16 / 7 〉 when t = 3 / 7, R ^ 2max = 16 / 7, at this time, the area has the maximum value, which is 16 / 7 π (PS: I don't want to

It is known that the equation of circle C is x ^ 2 + y ^ 2-2 (T + 3) x + 2 (1-4t ^ 2) y + 16t ^ 4 + 9 = 0 (t ∈ R) to find the orbit equation of the center of circle C

X ^ 2 + y ^ 2-2 (T + 3) x + 2 (1-4t ^ 2) y + 16t ^ 4 + 9 = 0 formula x ^ 2-2 (T + 3) x + (T + 3) ^ 2 + y ^ 2 + 2 (1-4t ^ 2) y + (1-4t ^ 2) ^ 2 = (T + 3) ^ 2 + (1-4t ^ 2) ^ 2, so the circle is (T + 3,4t ^ 2-1), the radius square = (T + 3) ^ 2 + (1-4t ^ 2) ^ 2 = 17T ^ 2-2t + 10 > x x x x x = t + 3, y = 1-4t ^ 2, TT = x-3, substitute y

If the chord length of the straight line L passing through the point m (- 3. - 3) is 10 cut by the circle X & # 178; + Y & # 178; + 4y-21 = 0, then the equation of the straight line L is

Equation: x-3y-6 = 0

The graph represented by equation x ^ 2 + y ^ 2 + 2x-4y-6 is

（x+1）²+（y-2）²=11
So the graph is a circle with (- 1, 2) as the center and R = √ 11
I wish you a happy study!