Equation of circle: x^2 +y^2 -4x -6y -3 = 0 Completing the squares: (x-2)^2 +(y-3)^2 = 16 square units Therefore center of circle is at (2, 3) and its radius is 4 units
Circle equation: x^2 -4x +y^2 -6y = 4 Completing the squares: (x-2)^2 +(y-3)^2 = 17 Point of contact: (6, 4) Center of circle: (2, 3) Slope of radius: 1/4 Slope of tangent line: -4 Tangent equation: y-4 = -4(x-6) => y = -4x+28 Tangent line equation in its general form: 4x+y-28 = 0
x2+y2+4x+6y-40 = 0 and x = 10+y Substitute the second equation into the first equation: (10+y)2+y2+4(10+y)+6y-40 = 0 2y2+30y+100 = 0 Divide all terms by 2: y2+15y+50 = 0 (y+10)(y+5) = 0 => y = -10 or y = -5 Substitute the above values into the second equation to find the points of intersection: Points of intersection are: (0, -10) and (5, -5)
If: 2x+y = 5 and x2-y2 = 3 Then the solutions work out as: (2, 1) and ( 14/3, -13/3)
x2 - 14x + 49 = (x - 7)2
y = 4x-3
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
No.
x^2+4x+7
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
y=-x^2+4x-3
Y = x2 + 4x + 11OK. That's very interesting.Is there a question ?
1 and 2
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
x(x+4) equals x2 plus 4x