I need solutions in exercise number 6 and 7 6. Determine the equation of the circle whose center is at (4, 5) and tangent to thecircle whose equation is x2 + y2 + 4x + 6y -23 = 0.A. x2 + y2 - 8x + 10y - 25 = 0B. x2 + y2 + 8x -10y + 25 0%3D(C) x2 + y2 - 8x - 10y + 25 = 0D. x2 + y2 - 8x - 10y - 25 = 0%3D7. A circle has a diameter whose ends are at (-3, 2) and (12, -6). Its Equation is:A. 4x2 + 4y2-36x +16y +192 = 0B) 4x2 + 4y2 - 36x +16y -192 = 0C. 4x2 + 4y2 -36x -16y - 192 = 0D. 4x2 + 4y2 - 36x +16y -192 = 08. The vertex of the parabola y2 -2x + 6y + 3 = 0 is at:%3D%3D%3DA. (-3, 3)В. (3,3)С. (3, -3)D.(-3, -3)9. Given the equation of the parabola: y2 - 8x – 4y - 20 = 0. The length of its latusrectum is:В. 4C. 6D. 810. What is the length of the latus rectum of the curve x2 = -12y?B.-3%3DА. 12С. 3D. -12then,-k)= (x-h)?seal benrth%3D

Name: I need solutions in exercise number 6 and 7 6. Determine the equation
Uploaded: 2024-03-20T07:07:26.000Z
Channel: Bartleby
Description: I need solutions in exercise number 6 and 7 6. Determine the equation of the circle whose center is at (4, 5) and tangent to thecircle whose equation is x2 + y2 + 4x + 6y -23 = 0.A. x2 + y2 - 8x + 10y - 25 = 0B. x2 + y2 + 8x -10y + 25 0%3D(C) x2 + y2