COORDINATE GEOMETRY 1 (With detailed solutions)

COORDINATE GEOMETRY 1

Question 1

(a) The end point coordinates of a line segment PQ are P(x₁, y₁) and Q(x₂, y₂). Prove that the coordinates of the point A(x, y) dividing the line segment PQ in the ratio m:n internally is:

A(x, y) = ( (mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n) )

(b) Show that for the point whose coordinates are given by x = 3cos t + 2, y = 3sin t - 4 is a circle.

Question 2

(a) Find the values of c such that the line x + y - c = 0 shall be a tangent to a circle x² + y² - 4x + 2 = 0. For each value of c, find the coordinates of the point of contact.

(b) Find the equation of the circle with center c(-1,2) and orthogonal to the circle x² + y² - 6x + 2y + 1 = 0.

(c) Find the equation of a circle passing through the points A(1,2), B(3,4) and orthogonal to a circle x² + y² + 8x + 2y - 22 = 0.

Question 3

(a) Find the equations of the tangent from the point (1,7) to the circle x² + y² = 5.

(b) Find the acute angle between the lines x + 4y = 12 and y - 2x + 6 = 0.

(c) A circle whose coordinate of center are both positive touches both axes of coordinate. If it also touches the line 3x - 4y + 6 = 0, find its equation.

Question 4

(a) Find the angle between the line y = 0 and the combined line equation (y/x)² = 3 with positive slope.

(b) The straight line 2y - x - 16 = 0 is perpendicular bisector of the line joining the points A and B. If A is at the point (-3, 4), find the coordinate of the point B.

(c) x² + y² - 16y + 32 = 0 and x² + y² - 18x + 2y + 32 = 0 are two circles. Show that the circles touch externally and find their point of contact.

Question 5

(a) Find the equation of the circle with center P(-1,2) and orthogonal to the circle x² + y² - 6x + 2y + 1 = 0.

(b) Find a pair of parallel lines from the equation 4x² + 4xy + y² - 6x - 3y - 4 = 0. Then, find the distance between them.

(c) Find the acute angle between the lines 3x - 2y - 8 = 0 and x - 5y + 7 = 0.

Coordinate Geometry 1 - Complete Solutions

Question 1

(a) Prove the section formula for internal division

Given points P(x₁,y₁) and Q(x₂,y₂), and point A(x,y) dividing PQ in ratio m:n

Using similar triangles:

(x - x₁)/(x₂ - x) = m/n

Cross-multiply: n(x - x₁) = m(x₂ - x)

Solve for x: nx - nx₁ = mx₂ - mx

x = (mx₂ + nx₁)/(m + n)

Similarly for y-coordinate:

y = (my₂ + ny₁)/(m + n)

Thus, A(x,y) = ( (mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n) )

(b) Show x = 3cos t + 2, y = 3sin t - 4 represents a circle

Rewrite equations:

x - 2 = 3cos t

y + 4 = 3sin t

Square and add both equations:

(x-2)² + (y+4)² = 9cos²t + 9sin²t = 9(cos²t + sin²t) = 9

This is the standard form of a circle:

(x - h)² + (y - k)² = r²

Final Answer: The equation represents a circle with center (2,-4) and radius 3

Question 2

(a) Find c such that x + y - c = 0 is tangent to x² + y² - 4x + 2 = 0

Rewrite circle equation in standard form:

x² - 4x + y² = -2

(x² - 4x + 4) + y² = -2 + 4

(x - 2)² + y² = 2

Center (2,0), radius r = √2

Condition for tangency: distance from center to line = radius

|2 + 0 - c|/√(1² + 1²) = √2

|2 - c|/√2 = √2

|2 - c| = 2

Solutions: 2 - c = 2 → c = 0

or 2 - c = -2 → c = 4

Final Answer: c = 0 or 4

Points of contact:

For c=0: Foot of perpendicular from (2,0) to x+y=0:

x = (2-0)/2 = 1, y = (0-0)/2 = 0 → (1,-1)

For c=4: Foot of perpendicular from (2,0) to x+y=4:

x = (2+4)/2 = 3, y = (0+4)/2 = 2 → (3,1)

(b) Find equation of circle with center (-1,2) orthogonal to x² + y² - 6x + 2y + 1 = 0

Given circle: Rewrite in standard form:

(x² - 6x) + (y² + 2y) = -1

(x - 3)² + (y + 1)² = 9

Center (3,-1), radius R = 3

Condition for orthogonality: d² = r₁² + r₂²

Distance between centers d = √[(3-(-1))² + (-1-2)²] = 5

Let r be radius of required circle: 5² = 3² + r² → r = 4

Final Answer: (x+1)² + (y-2)² = 16

(c) Find equation of circle through A(1,2), B(3,4) orthogonal to x² + y² + 8x + 2y - 22 = 0

Given circle: (x+4)² + (y+1)² = 39, center (-4,-1), R=√39

Let required circle: x² + y² + Dx + Ey + F = 0

Passes through A: 1 + 4 + D + 2E + F = 0 → D + 2E + F = -5

Passes through B: 9 + 16 + 3D + 4E + F = 0 → 3D + 4E + F = -25

Orthogonality condition: 2(-4)(D/2) + 2(-1)(E/2) = F - 22

-4D - E = F - 22 → 4D + E + F = 22

Solve system of equations:

D = -6, E = -2, F = 8

Final Answer: x² + y² - 6x - 2y + 8 = 0

Question 3

(a) Find equations of tangents from (1,7) to x² + y² = 5

Circle center (0,0), radius r = √5

Check point is outside: 1² + 7² = 50 > 5

General tangent equation: y = mx ± r√(1+m²)

Through (1,7): 7 = m(1) ± √5√(1+m²)

Square both sides: (7 - m)² = 5(1 + m²)

49 - 14m + m² = 5 + 5m² → 4m² + 14m - 44 = 0

2m² + 7m - 22 = 0 → m = [ -7 ± √(49 + 176) ]/4

m = (-7 ± 15)/4 → m = 2 or m = -5.5

Final Answers:
y = 2x ± √5√5 → y = 2x ± 5
y = -5.5x ± √5√31.25

(b) Find acute angle between x + 4y = 12 and y - 2x + 6 = 0

Slope of L₁: m₁ = -1/4

Slope of L₂: y = 2x - 6 → m₂ = 2

tanθ = |(m₂ - m₁)/(1 + m₁m₂)| = |(2 - (-1/4))/(1 + (-1/4)(2))| = |2.25/0.5| = 4.5

θ = tan⁻¹(4.5) ≈ 77.47°

Final Answer: Acute angle ≈ 77.47°

(c) Find equation of circle touching both axes and line 3x - 4y + 6 = 0

Center (a,a) since it touches both axes

Distance to line = radius a:

|3a - 4a + 6|/5 = a → |-a + 6| = 5a

Case 1: -a + 6 = 5a → 6 = 6a → a = 1

Case 2: -(-a + 6) = 5a → a - 6 = 5a → a = -1.5 (invalid, a > 0)

Final Answer: (x-1)² + (y-1)² = 1

Question 4

(a) Find angle between y = 0 and (y/x)² = 3 with positive slope

Equation simplifies to y² = 3x² → y = x√3 (positive slope)

Slope m = √3 → θ = tan⁻¹(√3) = 60°

Final Answer: Angle = 60°

(b) Find coordinates of B given A(-3,4) and perpendicular bisector 2y - x - 16 = 0

Midpoint M of AB lies on the line: M = ((x-3)/2, (y+4)/2)

2(y+4)/2 - (x-3)/2 - 16 = 0 → y + 4 - x/2 + 1.5 - 16 = 0

-x/2 + y - 10.5 = 0 → x - 2y + 21 = 0

Slope of AB: m₁ = (y-4)/(x+3)

Slope of bisector: m₂ = 1/2

Since perpendicular: m₁ × m₂ = -1 → (y-4)/(x+3) = -2

y - 4 = -2x - 6 → 2x + y + 2 = 0

Solve system: x - 2y = -21 and 2x + y = -2

x = -5, y = 8

Final Answer: B = (-5,8)

(c) Show circles touch externally and find point of contact

Circle 1: x² + (y-8)² = 32 → C₁(0,8), r₁ = √32 ≈ 5.66

Circle 2: (x-9)² + (y+1)² = 110 → C₂(9,-1), r₂ = √110 ≈ 10.49

Distance between centers:

d = √[(9-0)² + (-1-8)²] = √(81 + 81) = √162 ≈ 12.73

Check: d = r₁ + r₂ ≈ 5.66 + 10.49 ≈ 16.15 (Not matching)

Correction: The circles actually don't touch externally based on these calculations

Final Answer: The given circles do not touch externally

Question 5

(a) Find equation of circle with center (-1,2) orthogonal to x² + y² - 6x + 2y + 1 = 0

This is identical to Question 2(b) above

Final Answer: (x+1)² + (y-2)² = 16

(b) Find pair of parallel lines from 4x² + 4xy + y² - 6x - 3y - 4 = 0 and distance between them

Treat as quadratic in y: y² + (4x-3)y + (4x²-6x-4) = 0

For perfect square, discriminant must be zero:

(4x-3)² - 4(1)(4x²-6x-4) = 0

16x² - 24x + 9 - 16x² + 24x + 16 = 0 → 25 = 0 (No solution)

Alternative approach: Factor as (2x + y)² - 3(2x + y) - 4 = 0

Let z = 2x + y → z² - 3z - 4 = 0 → z = 4 or z = -1

Thus: 2x + y = 4 and 2x + y = -1

Distance between parallel lines:

d = |4 - (-1)|/√(2² + 1²) = 5/√5 = √5

Final Answers:
Parallel lines: 2x + y = 4 and 2x + y = -1
Distance: √5 units

(c) Find acute angle between 3x - 2y - 8 = 0 and x - 5y + 7 = 0

Slope of L₁: m₁ = 3/2

Slope of L₂: m₂ = 1/5

tanθ = |(m₁ - m₂)/(1 + m₁m₂)| = |(1.5 - 0.2)/(1 + 0.3)| ≈ 1.3/1.3 = 1

θ = tan⁻¹(1) = 45°

Final Answer: Acute angle = 45°