Question Statement

Find the points of intersection of the given line and the circle.

(i) $x - y + 1 = 0$ and $x^{2} + y^{2} - 3 x - 8 = 0$

(ii) $2 x + y + 45 = 0$ and $x^{2} + y^{2} - 2 x + 4 y - 11 = 0$

(iii) $3 x + 2 y + 1 = 0$ and $x^{2} + y^{2} - x + y + 2 = 0$

Background and Explanation

To find the intersection points of a line and a circle, we solve the system of equations simultaneously by substituting the linear equation into the circle equation. This yields a quadratic equation whose discriminant determines whether there are two distinct intersection points, one tangent point, or no real intersection.

Solution

Part (i)

We have the system: $x - y + 1 x^{2} + y^{2} - 3 x - 8 = 0 (i) = 0 (ii)$

For the points of intersection, we solve equations (i) and (ii) simultaneously.

From equation (i), we express $x$ in terms of $y$ : $x = y - 1$

Substituting this into equation (ii): $(y - 1)^{2} + y^{2} - 3 (y - 1) - 8 = 0$

Expanding and simplifying: $y^{2} - 2 y + 1 + y^{2} - 3 y + 3 - 8 = 0$ $2 y^{2} - 5 y - 4 = 0$

Using the quadratic formula $y = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ with $a = 2$ , $b = - 5$ , and $c = - 4$ : $y = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 2 ) ( - 4 )}{2 ( 2 )}$ $y = \frac{5 \pm 25 + 32}{4}$ $y = \frac{5 \pm 57}{4}$

This gives us two values for $y$ : $y = \frac{5 - 57}{4} and y = \frac{5 + 57}{4}$

Finding the corresponding $x$ -values:

Substituting $y = \frac{5 - 57}{4}$ back into equation (i): $x - (\frac{5 - 57}{4}) + 1 = 0$ $x = (\frac{5 - 57}{4}) - 1$ $x = \frac{5 - 57 - 4}{4}$ $x = \frac{1 - 57}{4}$

Substituting $y = \frac{5 + 57}{4}$ back into equation (i): $x - \frac{5 + 57}{4} + 1 = 0$ $x = \frac{5 + 57}{4} - 1$ $x = \frac{5 + 57 - 4}{4}$ $x = \frac{1 + 57}{4}$

Thus, the required points of intersection are: $(\frac{1 - 57}{4}, \frac{5 - 57}{4}) and (\frac{1 + 57}{4}, \frac{5 + 57}{4})$

Part (ii)

We have the system: $2 x + y + 45 x^{2} + y^{2} - 2 x + 4 y - 11 = 0 (i) = 0 (ii)$

For the point of intersection, we solve equations (i) and (ii) simultaneously.

From equation (i), we express $y$ in terms of $x$ : $y = - 2 x - 45$

Substituting this into equation (ii): $x^{2} + (- 2 x - 45)^{2} - 2 x + 4 (- 2 x - 45) - 11 = 0$

Expanding $(- 2 x - 45)^{2} = 4 x^{2} + 165 x + 80$ : $x^{2} + 4 x^{2} + 165 x + 80 - 2 x - 8 x - 165 - 11 = 0$

Combining like terms: $5 x^{2} + (165 - 10) x + (69 - 165) = 0$

Applying the quadratic formula: $x = \frac{- ( 16 5 - 10 ) \pm ( 16 5 - 10 ) ^{2} - 4 ( 5 ) ( 69 - 16 5 )}{2 ( 5 )}$

Calculating the discriminant: $(165 - 10)^{2} = 1280 - 3205 + 100 = 1380 - 3205$ $4 (5) (69 - 165) = 20 (69 - 165) = 1380 - 3205$

Thus: $x = \frac{- ( 16 5 - 10 ) \pm ( 1380 - 320 5 ) - ( 1380 - 320 5 )}{10}$ $x = \frac{- ( 16 5 - 10 ) \pm 0}{10}$ $x = \frac{10 - 16 5}{10} = \frac{5 - 8 5}{5}$

Since the discriminant is zero, there is exactly one solution for $x$ , indicating the line is tangent to the circle.

Substituting this $x$ -value back into equation (i) to find $y$ : $2 (\frac{5 - 8 5}{5}) + y + 45 = 0$ $y = - [\frac{2 ( 5 - 8 5 )}{5} + 45]$ $y = - [\frac{10 - 16 5 + 20 5}{5}]$ $y = - (\frac{10 + 4 5}{5}) = \frac{- 10 - 4 5}{5}$

Thus, $(\frac{5 - 8 5}{5}, \frac{- 10 + 4 5}{5})$ is the only point of intersection; that is, the line is tangent to the circle.

Part (iii)

We have the system: $3 x + 2 y + 1 x^{2} + y^{2} - x + y + 2 = 0 (i) = 0 (ii)$

For the points of intersection, we solve equations (i) and (ii) simultaneously.

From equation (i): $2 y = - 3 x - 1$ $y = \frac{- 3 x - 1}{2}$

Substituting into equation (ii): $x^{2} + (\frac{- 3 x - 1}{2})^{2} - x + \frac{- 3 x - 1}{2} + 2 = 0$

Expanding $(\frac{- 3 x - 1}{2})^{2} = \frac{9 x ^{2} + 6 x + 1}{4}$ : $x^{2} + \frac{9 x ^{2} + 6 x + 1}{4} - x + \frac{- 3 x - 1}{2} + 2 = 0$

Multiplying through by 4 to clear denominators: $4 x^{2} + 9 x^{2} + 6 x + 1 - 4 x - 6 x - 2 + 8 = 0$ $13 x^{2} - 4 x + 7 = 0$

Applying the quadratic formula: $x = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( 13 ) ( 7 )}{2 ( 13 )}$ $x = \frac{4 \pm 16 - 364}{26}$ $x = \frac{4 \pm - 348}{26}$ $x = \frac{4 \pm i 348}{26}$

This gives us two complex solutions: $x = \frac{4 - i 348}{26} and x = \frac{4 + i 348}{26}$

Finding the corresponding $y$ -values:

For $x = \frac{4 - i 348}{26}$ , substituting into equation (i): $3 (\frac{4 - i 348}{26}) + 2 y + 1 = 0$ $2 y + \frac{12 - 3 i 348}{26} + 1 = 0$ $2 y + \frac{12 - 3 i 348 + 26}{26} = 0$ $2 y + \frac{38 - 3 i 348}{26} = 0$ $2 y = - \frac{38 - 3 i 348}{26}$ $y = - \frac{38 - 3 i 348}{52}$

For $x = \frac{4 + i 348}{26}$ , substituting into equation (i): $3 (\frac{4 + i 348}{26}) + 2 y + 1 = 0$ $2 y + \frac{12 + 3 i 348}{26} + 1 = 0$ $2 y + \frac{12 + 3 i 348 + 26}{26} = 0$ $2 y + \frac{38 + 3 i 348}{26} = 0$ $2 y = - \frac{38 + 3 i 348}{26}$ $y = - \frac{38 + 3 i 348}{52}$

Thus, the points of intersection are: $(\frac{4 - i 348}{26}, - \frac{38 - 3 i 348}{52}) and (\frac{4 + i 348}{26}, - \frac{38 + 3 i 348}{52})$

These complex points indicate that the line does not intersect the circle in the real plane.

Key Formulas or Methods Used

Substitution method: Expressing one variable from the linear equation and substituting into the circle equation
Quadratic formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ for solving $a x^{2} + b x + c = 0$
Discriminant analysis: $Δ = b^{2} - 4 a c$ determines the nature of roots:
- $Δ > 0$ : Two distinct real intersection points (secant line)
- $Δ = 0$ : One real intersection point (tangent line)
- $Δ < 0$ : No real intersection points (line misses the circle)

Summary of Steps

Isolate one variable from the linear equation (e.g., $x = \dots$ or $y = \dots$ )
Substitute this expression into the circle equation to eliminate one variable
Simplify the resulting equation to standard quadratic form $a x^{2} + b x + c = 0$ (or in terms of $y$ )
Calculate the discriminant $Δ = b^{2} - 4 a c$ to determine the nature of intersection
Solve the quadratic using the quadratic formula to find the coordinate values
Back-substitute these values into the linear equation to find the corresponding coordinates of the intersection points