1
Obtain the equation of the circles with radius \( 5 \, \text{cm} \) and touching \( x \)-axis at the origin in general form.
1 Circle touches x-axis at (0,0) with radius 5cm.
2 Center must be at (0,5) or (0,-5).
3 Standard form equations:
\( x^2 + (y-5)^2 = 25 \) or \( x^2 + (y+5)^2 = 25 \)
Final Answer: \( x^2 + y^2 \pm 10y = 0 \)
2
Find the equation of the circle with centre (2, -1) and passing through (3, 6) in standard form.
1 Center at (2,-1), point on circle at (3,6).
2 Calculate radius:
\( r = \sqrt{(3-2)^2 + (6-(-1))^2} = \sqrt{1 + 49} = 5\sqrt{2} \)
Final Answer: \( (x-2)^2 + (y+1)^2 = 50 \)
3
Find the equation of circles that touch both axes and pass through (-4, -2) in general form.
1 Center must be at (±r, ±r).
2 For (-4,-2), center is (-r,-r).
3 Solve \( (-4+r)^2 + (-2+r)^2 = r^2 \)
\( r^2 - 12r + 20 = 0 \) → r = 10 or r = 2
Final Answers:
\( x^2 + y^2 + 20x + 20y + 100 = 0 \) (r=10)
\( x^2 + y^2 + 4x + 4y + 4 = 0 \) (r=2)
4
Find the equation of the circle with centre (2, 3) passing through the intersection of \(3x - 2y - 1 = 0\) and \(4x + y - 27 = 0\).
1 Find intersection point:
Solve \( 3x - 2y = 1 \) and \( 4x + y = 27 \)
Solution: (5,7)
2 Calculate radius:
\( r = \sqrt{(5-2)^2 + (7-3)^2} = 5 \)
Final Answer: \( (x-2)^2 + (y-3)^2 = 25 \)
5
Obtain the equation of the circle for which (3, 4) and (2, -7) are the ends of a diameter.
1 Find center (midpoint):
\( \left( \frac{3+2}{2}, \frac{4+(-7)}{2} \right) = (2.5, -1.5) \)
2 Calculate radius (half diameter length):
\( \frac{\sqrt{(3-2)^2 + (4-(-7))^2}}{2} = \frac{\sqrt{122}}{2} \)
Final Answer: \( (x-2.5)^2 + (y+1.5)^2 = 30.5 \)
6
Find the equation of the circle through (1, 0), (-1, 0), and (0, 1).
1 Let equation be \( x^2 + y^2 + Dx + Ey + F = 0 \)
2 Substitute points:
(1,0): \( 1 + D + F = 0 \)
(-1,0): \( 1 - D + F = 0 \)
(0,1): \( 1 + E + F = 0 \)
3 Solve system: D = 0, F = -1, E = 0
Final Answer: \( x^2 + y^2 = 1 \)
7
A circle of area \(9\pi\) has two diameters along \(x + y = 5\) and \(x - y = 1\). Find its equation.
1 Find center at intersection of diameters:
Solve \( x + y = 5 \) and \( x - y = 1 \)
Center at (3,2)
2 Calculate radius:
\( \pi r^2 = 9\pi \) ⇒ \( r = 3 \)
Final Answer: \( (x-3)^2 + (y-2)^2 = 9 \)
8
If \( y = 2\sqrt{2}x + c \) is tangent to \( x^2 + y^2 = 16 \), find \( c \).
1 Condition for tangency: distance from center (0,0) to line equals radius (4).
2 Distance formula:
\( \frac{|2\sqrt{2}(0) - 1(0) + c|}{\sqrt{(2\sqrt{2})^2 + (-1)^2}} = 4 \)
\( \frac{|c|}{3} = 4 \) ⇒ \( c = \pm 12 \)
Final Answer: \( c = 12 \) or \( c = -12 \)
9
Find the tangent and normal to \( x^2 + y^2 - 6x + 6y - 8 = 0 \) at (2, 2).
1 Rewrite circle equation:
\( (x-3)^2 + (y+3)^2 = 26 \)
Center at (3,-3)
2 Slope of normal (center to point):
\( m = \frac{2-(-3)}{2-3} = -5 \)
3 Equations:
Normal: \( y-2 = -5(x-2) \) or \( 5x + y - 12 = 0 \)
Tangent: \( y-2 = \frac{1}{5}(x-2) \) or \( x - 5y + 8 = 0 \)
Final Answers:
Tangent: \( x - 5y + 8 = 0 \)
Normal: \( 5x + y - 12 = 0 \)
10
Determine whether (-2,1), (0,0), and (-4,-3) lie outside, on or inside \( x^2 + y^2 - 5x + 2y - 5 = 0 \).
1 Rewrite circle equation:
\( (x-2.5)^2 + (y+1)^2 = 12.25 \)
Center at (2.5,-1), r = 3.5
2 Calculate distance for each point:
(-2,1): \( \sqrt{(-2-2.5)^2 + (1-(-1))^2} = \sqrt{24.25} > 3.5 \) → Outside
(0,0): \( \sqrt{(0-2.5)^2 + (0-(-1))^2} = \sqrt{7.25} < 3.5 \) → Inside
(-4,-3): \( \sqrt{(-4-2.5)^2 + (-3-(-1))^2} = \sqrt{46.25} > 3.5 \) → Outside
Final Answer:
(-2,1) → Outside
(0,0) → Inside
(-4,-3) → Outside
11
Find center and radius of:
(i) \(x^2 + (y + 2)^2 = 0\)
(ii) \(x^2 + y^2 + 6x - 4y + 4 = 0\)
(iii) \(x^2 + y^2 - x + 2y - 3 = 0\)
(iv) \(2x^2 + 2y^2 - 6x + 4y + 2 = 0\)
1 (i) \(x^2 + (y + 2)^2 = 0\):
Center: (0,-2), Radius: 0 (degenerate circle)
2 (ii) Complete squares:
\( (x+3)^2 + (y-2)^2 = 9 \)
Center: (-3,2), Radius: 3
3 (iii) Complete squares:
\( (x-0.5)^2 + (y+1)^2 = 4.25 \)
Center: (0.5,-1), Radius: √4.25
4 (iv) Divide by 2 first:
\( (x-1.5)^2 + (y+1)^2 = 1.5 \)
Center: (1.5,-1), Radius: √1.5
12
If \(3x^2 + (3 - p)xy + qy^2 - 2px = 8pq\) represents a circle, find \(p\) and \(q\).
1 Conditions for circle:
(1) Coefficient of \(x^2\) = coefficient of \(y^2\) ⇒ \(3 = q\)
(2) No xy term ⇒ \(3 - p = 0\) ⇒ \(p = 3\)
2 Verify with p=3, q=3:
\(3x^2 + 3y^2 - 6x = 72\) ⇒ \(x^2 + y^2 - 2x - 24 = 0\)
3 Find center and radius:
\( (x-1)^2 + y^2 = 25 \)
Center: (1,0), Radius: 5
Final Answer: \( p = 3 \), \( q = 3 \)