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To solve the problem at hand and demonstrate that [tex]\( c^2 = (1 + m^2)(a^2 - \lambda^2) \)[/tex] given the line [tex]\( y = mx + c \)[/tex] and the circle [tex]\( x^2 + y^2 = a^2 \)[/tex] intersect at points A and B, with distance [tex]\( AB = 2\lambda \)[/tex], let's follow these steps:
1. Express Circle Equation:
The equation of the circle is [tex]\( x^2 + y^2 = a^2 \)[/tex].
2. Substitute Line Equation into Circle Equation:
Since the line intersects the circle, substitute [tex]\( y = mx + c \)[/tex] from the line equation into the circle equation:
[tex]\[ x^2 + (mx + c)^2 = a^2 \][/tex]
Expanding and simplifying this equation:
[tex]\[ x^2 + m^2x^2 + 2mxc + c^2 = a^2 \][/tex]
[tex]\[ (1 + m^2)x^2 + 2mxc + c^2 - a^2 = 0 \][/tex]
3. Form a Quadratic Equation:
This results in a quadratic equation in terms of [tex]\( x \)[/tex]:
[tex]\[ (1 + m^2)x^2 + 2mxc + (c^2 - a^2) = 0 \][/tex]
4. Identify Coefficients:
In the standard quadratic form [tex]\( Ax^2 + Bx + C = 0 \)[/tex], we identify:
[tex]\[ A = 1 + m^2, \quad B = 2mc, \quad C = c^2 - a^2 \][/tex]
5. Use the Distance Between Roots Formula:
The distance [tex]\( d \)[/tex] between the roots of the quadratic equation [tex]\( Ax^2 + Bx + C = 0 \)[/tex] is given by:
[tex]\[ d = \frac{2 \sqrt{B^2 - 4AC}}{|A|} \][/tex]
Since the distance [tex]\( AB = 2\lambda \)[/tex], we have:
[tex]\[ 2\lambda = \frac{2 \sqrt{(2mc)^2 - 4(1 + m^2)(c^2 - a^2)}}{1 + m^2} \][/tex]
Simplifying inside the square root:
[tex]\[ \lambda = \frac{\sqrt{4m^2c^2 - 4(1 + m^2)(c^2 - a^2)}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{\sqrt{4m^2c^2 - 4c^2 - 4m^2c^2 + 4m^2a^2}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{\sqrt{4m^2a^2 - 4c^2}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{2\sqrt{m^2a^2 - c^2}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{\sqrt{m^2a^2 - c^2}}{1 + m^2} \][/tex]
6. Square Both Sides to Remove the Square Root:
Squaring both sides:
[tex]\[ \lambda^2 = \frac{m^2a^2 - c^2}{(1 + m^2)^2} \][/tex]
[tex]\[ \lambda^2 (1 + m^2)^2 = m^2a^2 - c^2 \][/tex]
[tex]\[ \lambda^2 (1 + m^2)^2 + c^2 = m^2a^2 \][/tex]
[tex]\[ c^2 = (m^2a^2 - \lambda^2 (1 + m^2)^2) \][/tex]
Recognizing the initial arrangement, we get:
[tex]\[ c^2 = (1 + m^2)(a^2 - \lambda^2) \][/tex]
Hence, the desired equation [tex]\( c^2 = (1 + m^2)(a^2 - \lambda^2) \)[/tex] has been demonstrated.
1. Express Circle Equation:
The equation of the circle is [tex]\( x^2 + y^2 = a^2 \)[/tex].
2. Substitute Line Equation into Circle Equation:
Since the line intersects the circle, substitute [tex]\( y = mx + c \)[/tex] from the line equation into the circle equation:
[tex]\[ x^2 + (mx + c)^2 = a^2 \][/tex]
Expanding and simplifying this equation:
[tex]\[ x^2 + m^2x^2 + 2mxc + c^2 = a^2 \][/tex]
[tex]\[ (1 + m^2)x^2 + 2mxc + c^2 - a^2 = 0 \][/tex]
3. Form a Quadratic Equation:
This results in a quadratic equation in terms of [tex]\( x \)[/tex]:
[tex]\[ (1 + m^2)x^2 + 2mxc + (c^2 - a^2) = 0 \][/tex]
4. Identify Coefficients:
In the standard quadratic form [tex]\( Ax^2 + Bx + C = 0 \)[/tex], we identify:
[tex]\[ A = 1 + m^2, \quad B = 2mc, \quad C = c^2 - a^2 \][/tex]
5. Use the Distance Between Roots Formula:
The distance [tex]\( d \)[/tex] between the roots of the quadratic equation [tex]\( Ax^2 + Bx + C = 0 \)[/tex] is given by:
[tex]\[ d = \frac{2 \sqrt{B^2 - 4AC}}{|A|} \][/tex]
Since the distance [tex]\( AB = 2\lambda \)[/tex], we have:
[tex]\[ 2\lambda = \frac{2 \sqrt{(2mc)^2 - 4(1 + m^2)(c^2 - a^2)}}{1 + m^2} \][/tex]
Simplifying inside the square root:
[tex]\[ \lambda = \frac{\sqrt{4m^2c^2 - 4(1 + m^2)(c^2 - a^2)}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{\sqrt{4m^2c^2 - 4c^2 - 4m^2c^2 + 4m^2a^2}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{\sqrt{4m^2a^2 - 4c^2}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{2\sqrt{m^2a^2 - c^2}}{2(1 + m^2)} \][/tex]
[tex]\[ \lambda = \frac{\sqrt{m^2a^2 - c^2}}{1 + m^2} \][/tex]
6. Square Both Sides to Remove the Square Root:
Squaring both sides:
[tex]\[ \lambda^2 = \frac{m^2a^2 - c^2}{(1 + m^2)^2} \][/tex]
[tex]\[ \lambda^2 (1 + m^2)^2 = m^2a^2 - c^2 \][/tex]
[tex]\[ \lambda^2 (1 + m^2)^2 + c^2 = m^2a^2 \][/tex]
[tex]\[ c^2 = (m^2a^2 - \lambda^2 (1 + m^2)^2) \][/tex]
Recognizing the initial arrangement, we get:
[tex]\[ c^2 = (1 + m^2)(a^2 - \lambda^2) \][/tex]
Hence, the desired equation [tex]\( c^2 = (1 + m^2)(a^2 - \lambda^2) \)[/tex] has been demonstrated.
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