Final answer:
a. The x-intercepts are -7 and 3. b. The equations of the lines tangent to the circle at the x-intercepts are y = (12/(ya^3))(x+7) - 2 and y = (-5/(ya^3))(x-3) - 2. c. The equation of the circle with center A that passes through point O is (x-3)^2 + (y+2)^2 = 104. d. The intersection points of circle k (O) and circle k (A) are given by the equation 10x - 4y - ya^6 - y^2 = 75.
Step-by-step explanation:
a. To find the x-intercepts, we set y=0 and solve for x. Plugging in y=0 in the equation of the circle, we get (x+2)^2 + (ya^3)^2 = 25, (x+2)^2 = 25, x+2 = ±√25, x = -2±5 = -7 and 3. So the x-intercepts are -7 and 3.
b. To find the equations of the lines tangent to the circle at the x-intercepts, we need to find the derivative of the equation of the circle. Differentiating (x+2)^2 + (ya^3)^2 = 25 with respect to x, we get 2(x+2) + 2ya^3(y') = 0. Plugging in the x-intercepts, we have at x=-7: -12 + 2ya^3(y') = 0, ya^3(y') = 12, and at x=3: 10 + 2ya^3(y') = 0, ya^3(y') = -5. Therefore, the equations of the lines tangent to the circle at the x-intercepts are y = (12/(ya^3))(x+7) - 2 and y = (-5/(ya^3))(x-3) - 2.
c. To find the equation of a circle with center A that passes through point O, we first need to find the coordinates of point A. Since A is the intersection of the two tangent lines, we can set their equations equal to each other: (12/(ya^3))(x+7) - 2 = (-5/(ya^3))(x-3) - 2. Solving for x, we get x = 3. Substituting x=3 into one of the tangent lines, we can solve for y: y = (-5/(ya^3))(3-3) - 2. Simplifying, we obtain y = -2. Therefore, the coordinates of point A are (3, -2). Now we can find the equation of the circle with center A that passes through point O. Using the distance formula, the equation is (x-3)^2 + (y+2)^2 = OA^2, where OA is the distance between O and A. We can find OA by substituting the coordinates of O and A into the distance formula: OA = √((-7-3)^2 + (0+2)^2). Simplifying, we get OA = √(100 + 4) = √104. Therefore, the equation of the circle is (x-3)^2 + (y+2)^2 = 104.
d. To find the intersection points of circle k (O) and circle k (A), we need to solve the system of equations formed by setting the equations of the two circles equal to each other: (x+2)^2 + (ya^3)^2 = 25 and (x-3)^2 + (y+2)^2 = 104. Simplifying these equations, we get x^2 + 4x + 4 + ya^6 = 25 and x^2 - 6x + 9 + y^2 + 4y + 4 = 104. Combining like terms and subtracting one equation from the other, we get 10x - 4y - ya^6 - y^2 = 75. This equation represents the intersection points between the circles k (O) and k (A).