The coordinates of the centre of a circle is (-2, 3). If its area is 25πcm2, find its equation.
x2+y2−4x−6y−12=0
x2+y2−4x+6y−12=0
x2+y2+4x+6y−12=0
x2+y2+4x−6y−12=0
Correct answer is D
Equation of a circle with centre coordinates (a, b) : (x−a)2+(y−b)2=r2
Area of circle = πr2=25πcm2⟹r2=25
∴
(a, b) = (-2, 3)
Equation: (x - (-2))^{2} + (y - 3)^{2} = 5^{2}
x^{2} + 4x + 4 + y^{2} - 6y + 9 = 25 \implies x^{2} + y^{2} + 4x - 6y + 13 - 25 = 0
= x^{2} + y^{2} + 4x - 6y - 12 = 0
-4
-1
3
4
Correct answer is C
y = 1 - 3x + 2x^{3}
\frac{\mathrm d y}{\mathrm d x} = -3 + 6x^{2}
At (1, 0), \frac{\mathrm d y}{\mathrm d x} = -3 + 6(1^{2}) = -3 + 6 = 3
y = mx - 3 \implies \frac{\mathrm d y}{\mathrm d x} = m = 3 (Tangent with equal gradient)
-1
-2
-3
-4
Correct answer is D
x + 3 = 0 \implies x = -3
Using remainder theorem, if x + 3 is a factor, f(-3) = 0.
f(-3) = (-3)^{3} + 3(-3)^{2} + n(-3) - 12 = 0
-27 + 27 - 3n - 12 = 0 \implies -3n = 12
n = -4
x < -4 or x > 2
x < -2 or x > 4
-2 < x < 4
-4 < x < 2
Correct answer is B
No explanation has been provided for this answer.
What is the angle between a = (3i - 4j) and b = (6i + 4j)?
13°
87°
100°
110°
Correct answer is B
a . b = |a||b| \cos \theta
a = 3i - 4j; b = 6i + 4j
18 - 16 = (\sqrt{3^{2} + (-4)^{2}})(\sqrt{6^{2} + 4^{2}}) \cos \theta
2 = 5\sqrt{52} \cos \theta
\cos \theta = \frac{2}{5\sqrt{52}} = 0.0555
\theta = 86.8° \approxeq 87°