A circle and a line have the equations \(x^2+y^2+6 x-8 y=0\) and \(y=m x-\frac{1}{3}\) respectively. Find the values or the range of values of \(m\) for which the line
(i) intersects the circle at two distinct points,
(ii) is a tangent to the circle,
(iii) does not meet the circle.
- Write the circle and line equations clearly:
\( \text{Circle: } x^2 + y^2 + 6x – 8y = 0, \quad \text{Line: } y = m x – \frac{1}{3}\).
- Substitute y from the line into the circle:
\(x^2 + \bigl(m x – \frac{1}{3}\bigr)^2 + 6x – 8\bigl(m x – \frac{1}{3}\bigr) = 0\)
- Simplify to form a quadratic in \(x^2\):
\(x^2 + \bigl(m^2 x^2 – \frac{2m}{3}x + \frac{1}{9}\bigr) + 6x – 8m x + \frac{8}{3} = 0\).
Combine like terms carefully:
- \(x^2\) and \(m^2 x^2\): \((1 + m^2)x^2\).
- \(x\) terms: \(6x – 8mx – \frac{2m}{3}x = \bigl(6 – 8m – \frac{2m}{3}\bigr)x\).
- Constant terms: \(\frac{1}{9} + \frac{8}{3} = \frac{1}{9} + \frac{24}{9} = \frac{25}{9}\).
Hence the quadratic in \(x^2\) is \((1 + m^2)x^2 + \Bigl(6 – 8m – \frac{2m}{3}\Bigr)x + \frac{25}{9} = 0\).
It is often easier to clear denominators; multiplying through by 9 gives \(9(1 + m^2)x^2 + \bigl(54 – 78m\bigr)x + 25 = 0\).
- Use the discriminant Δ for intersection conditions:
A quadratic \(ax^2 + bx + c = 0 \) has
- two distinct real solutions if \(\Delta = b^2 – 4ac > 0\),
- one (repeated) real solution (tangent) if \(\Delta = 0\),
- no real solutions if \(\Delta < 0\).
Here, \(a = 9(1 + m^2) \),\( b = 54 – 78m \), \(c = 25\). So \(\Delta = (54 – 78m)^2 \;-\; 4 \cdot 9(1 + m^2)\cdot 25 \).
After simplification/factorization, one obtains \( \Delta = 72\bigl(24m – 7\bigr)\bigl(3m – 4\bigr)\).
- Analyze the sign of Δ\Delta:
Since \(72 > 0\), the sign of \(\Delta\) depends on the product \((24m – 7)(3m – 4)\).
- Zeros of the factors:
- \(24m – 7 = 0 , m = \frac{7}{24} \).
- \( 3m – 4 = 0 , m = \frac{4}{3} \).
Order these critical values on the number line: \(\frac{7}{24}\approx 0.2917\) and \(\frac{4}{3}\approx 1.3333 \).
- For \( m < \frac{7}{24} \) : both \((24m-7)\) and \((3m−4)\)are negative, so their product is positive \( \Delta>0 \).
- For \(\frac{7}{24} < m < \frac{4}{3}\): one factor is positive and the other negative, so their product is negative \(\Delta<0\).
- For \(m > \frac{4}{3}\): both factors are positive, so their product is positive \( \Delta>0 \).
- Conclusion by cases:
- Two distinct intersections \( \Delta>0 \)
\(m < \frac{7}{24} \quad \text{or}\quad m > \frac{4}{3} \).
- Tangent to the circle \( \Delta=0 \)
\( m = \frac{7}{24} \quad \text{or}\quad m = \frac{4}{3} \).
- No real intersection \(\Delta<0 \)
\(\frac{7}{24} < m < \frac{4}{3} \).