Question Statement

The equation of a circle is $x^{2} + y^{2} - 4 x + y - 7 = 0$ and the equation of a line is $2 x - y + c = 0$ . Find the value(s) of $c$ such that the line:

(i) Intersects the circle at two distinct points.

(ii) Is tangent to the circle.

(iii) Has no common point with the circle.

Background and Explanation

To determine how a line intersects a circle, we substitute the linear equation into the circle's equation to obtain a quadratic equation in one variable. The discriminant of this quadratic tells us whether there are two solutions (secant), one solution (tangent), or no real solutions (the line misses the circle).

Solution

First, express $y$ from the line equation: $y = 2 x + c$

Substitute this into the circle equation $x^{2} + y^{2} - 4 x + y - 7 = 0$ : $x^{2} + (2 x + c)^{2} - 4 x + (2 x + c) - 7 = 0$

Expand $(2 x + c)^{2}$ and simplify: $x^{2} + 4 x^{2} + 4 c x + c^{2} - 4 x + 2 x + c - 7 = 0$ $5 x^{2} + (4 c - 2) x + (c^{2} + c - 7) = 0$

This is a quadratic equation in $x$ . For the line and circle to intersect, this quadratic must have real solutions. We analyze its discriminant:

$Discriminant Δ = (4 c - 2)^{2} - 4 (5) (c^{2} + c - 7)$

Calculate step-by-step: $Δ = 4 (2 c - 1)^{2} - 20 (c^{2} + c - 7) = 4 (4 c^{2} - 4 c + 1) - 20 c^{2} - 20 c + 140 = 16 c^{2} - 16 c + 4 - 20 c^{2} - 20 c + 140 = - 4 c^{2} - 36 c + 144 = - 4 (c^{2} + 9 c - 36) = - 4 (c^{2} + 12 c - 3 c - 36) = - 4 [c (c + 12) - 3 (c + 12)] = - 4 (c - 3) (c + 12)$

Thus, $Δ = - 4 (c - 3) (c + 12)$ .

Part (i): Two distinct intersection points ( $Δ > 0$ )

For two distinct points, we require: $- 4 (c - 3) (c + 12) > 0$

Dividing both sides by $- 4$ (and reversing the inequality): $(c - 3) (c + 12) < 0$

The product of two numbers is negative when one is positive and the other is negative.

Case 1: $c - 3 > 0$ and $c + 12 < 0$
$\Rightarrow c > 3$ and $c < - 12$ (impossible, no such number exists)
Case 2: $c - 3 < 0$ and $c + 12 > 0$
$\Rightarrow c < 3$ and $c > - 12$

Therefore: $- 12 < c < 3$

Part (ii): Tangent to the circle ( $Δ = 0$ )

For the line to be tangent (touching at exactly one point): $- 4 (c - 3) (c + 12) = 0$ $(c - 3) (c + 12) = 0$

This gives: $c = 3 or c = - 12$

Part (iii): No common points ( $Δ < 0$ )

For no intersection: $- 4 (c - 3) (c + 12) < 0$

Dividing by $- 4$ (reversing the inequality): $(c - 3) (c + 12) > 0$

The product is positive when both factors are positive or both are negative.

Case 1: $c - 3 > 0$ and $c + 12 > 0$
$\Rightarrow c > 3$ and $c > - 12$
Both conditions are satisfied when $c > 3$ (i.e., $3 < c < \infty$ )
Case 2: $c - 3 < 0$ and $c + 12 < 0$
$\Rightarrow c < 3$ and $c < - 12$
Both conditions are satisfied when $c < - 12$ (i.e., $- \infty < c < - 12$ )

Therefore, the line has no common points with the circle when: $c < - 12 or c > 3$

Key Formulas or Methods Used

Substitution method: Eliminating $y$ by substituting the line equation into the circle equation
Quadratic discriminant: For $a x^{2} + b x + c = 0$ , the discriminant is $Δ = b^{2} - 4 a c$
Intersection conditions:
- $Δ > 0$ : Line is a secant (two distinct points)
- $Δ = 0$ : Line is tangent (one point)
- $Δ < 0$ : Line does not intersect the circle
Inequality analysis: Sign analysis of quadratic expressions using critical points

Summary of Steps

Substitute $y = 2 x + c$ from the line into the circle equation $x^{2} + y^{2} - 4 x + y - 7 = 0$
Simplify to obtain the quadratic: $5 x^{2} + (4 c - 2) x + (c^{2} + c - 7) = 0$
Calculate the discriminant: $Δ = - 4 (c - 3) (c + 12)$
For two distinct points: Set $Δ > 0$ , giving $- 12 < c < 3$
For tangent: Set $Δ = 0$ , giving $c = 3$ or $c = - 12$
For no intersection: Set $Δ < 0$ , giving $c < - 12$ or $c > 3$

Question Statement

The equation of a circle is $x^{2} + y^{2} - 4 x + y - 7 = 0$ and the equation of a line is $2 x - y + c = 0$ . Find the value(s) of $c$ such that the line:

(i) Intersects the circle at two distinct points.

(ii) Is tangent to the circle.

(iii) Has no common point with the circle.

Background and Explanation

Solution

First, express $y$ from the line equation: $y = 2 x + c$

Substitute this into the circle equation $x^{2} + y^{2} - 4 x + y - 7 = 0$ : $x^{2} + (2 x + c)^{2} - 4 x + (2 x + c) - 7 = 0$

Expand $(2 x + c)^{2}$ and simplify: $x^{2} + 4 x^{2} + 4 c x + c^{2} - 4 x + 2 x + c - 7 = 0$ $5 x^{2} + (4 c - 2) x + (c^{2} + c - 7) = 0$

This is a quadratic equation in $x$ . For the line and circle to intersect, this quadratic must have real solutions. We analyze its discriminant:

$Discriminant Δ = (4 c - 2)^{2} - 4 (5) (c^{2} + c - 7)$

Thus, $Δ = - 4 (c - 3) (c + 12)$ .

Part (i): Two distinct intersection points ( $Δ > 0$ )

For two distinct points, we require: $- 4 (c - 3) (c + 12) > 0$

Dividing both sides by $- 4$ (and reversing the inequality): $(c - 3) (c + 12) < 0$

The product of two numbers is negative when one is positive and the other is negative.

Case 1: $c - 3 > 0$ and $c + 12 < 0$
$\Rightarrow c > 3$ and $c < - 12$ (impossible, no such number exists)
Case 2: $c - 3 < 0$ and $c + 12 > 0$
$\Rightarrow c < 3$ and $c > - 12$

Therefore: $- 12 < c < 3$

Part (ii): Tangent to the circle ( $Δ = 0$ )

For the line to be tangent (touching at exactly one point): $- 4 (c - 3) (c + 12) = 0$ $(c - 3) (c + 12) = 0$

This gives: $c = 3 or c = - 12$

Part (iii): No common points ( $Δ < 0$ )

For no intersection: $- 4 (c - 3) (c + 12) < 0$

Dividing by $- 4$ (reversing the inequality): $(c - 3) (c + 12) > 0$

The product is positive when both factors are positive or both are negative.

Case 1: $c - 3 > 0$ and $c + 12 > 0$
$\Rightarrow c > 3$ and $c > - 12$
Both conditions are satisfied when $c > 3$ (i.e., $3 < c < \infty$ )
Case 2: $c - 3 < 0$ and $c + 12 < 0$
$\Rightarrow c < 3$ and $c < - 12$
Both conditions are satisfied when $c < - 12$ (i.e., $- \infty < c < - 12$ )

Therefore, the line has no common points with the circle when: $c < - 12 or c > 3$

Key Formulas or Methods Used

Substitution method: Eliminating $y$ by substituting the line equation into the circle equation
Quadratic discriminant: For $a x^{2} + b x + c = 0$ , the discriminant is $Δ = b^{2} - 4 a c$
Intersection conditions:
- $Δ > 0$ : Line is a secant (two distinct points)
- $Δ = 0$ : Line is tangent (one point)
- $Δ < 0$ : Line does not intersect the circle
Inequality analysis: Sign analysis of quadratic expressions using critical points

Summary of Steps

Substitute $y = 2 x + c$ from the line into the circle equation $x^{2} + y^{2} - 4 x + y - 7 = 0$
Simplify to obtain the quadratic: $5 x^{2} + (4 c - 2) x + (c^{2} + c - 7) = 0$
Calculate the discriminant: $Δ = - 4 (c - 3) (c + 12)$
For two distinct points: Set $Δ > 0$ , giving $- 12 < c < 3$
For tangent: Set $Δ = 0$ , giving $c = 3$ or $c = - 12$
For no intersection: Set $Δ < 0$ , giving $c < - 12$ or $c > 3$

7.2 Q-2

Question Statement

Background and Explanation

Solution

Part (i): Two distinct intersection points ( $Δ > 0$ )

Part (ii): Tangent to the circle ( $Δ = 0$ )

Part (iii): No common points ( $Δ < 0$ )

Key Formulas or Methods Used

Summary of Steps

7.2 Q-2

Question Statement

Background and Explanation

Solution

Part (i): Two distinct intersection points ( $Δ > 0$ )

Part (ii): Tangent to the circle ( $Δ = 0$ )

Part (iii): No common points ( $Δ < 0$ )

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Part (i): Two distinct intersection points (Δ>0)

Part (ii): Tangent to the circle (Δ=0)

Part (iii): No common points (Δ<0)

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Part (i): Two distinct intersection points (Δ>0)

Part (ii): Tangent to the circle (Δ=0)

Part (iii): No common points (Δ<0)

Key Formulas or Methods Used

Summary of Steps

Part (i): Two distinct intersection points ( $Δ > 0$ )

Part (ii): Tangent to the circle ( $Δ = 0$ )

Part (iii): No common points ( $Δ < 0$ )

Part (i): Two distinct intersection points ( $Δ > 0$ )

Part (ii): Tangent to the circle ( $Δ = 0$ )

Part (iii): No common points ( $Δ < 0$ )