if y=mx+c and x²+y²=a² i)intersects at A and B ii)AB=2∆ then show that c²=(1+m²)(a²-∆²)
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See the image above. x^2 + y^2 =a^2 is a circle with center A(0,0) and radius a.
y = mx + c is equation for straight line. It touches the x axis at C(-c/m,0) and y axis at B(0,c).
For the line to touch the circle, it has to be a tangent as shown above (line BC).
From the figure, Area of Triangle ABC = 1/2 * AC * AB = 1/2 * BC * AD
BC^2 = AB^2 + AC^2 (Pythagoras Theorem)
So, squaring both sides, we have
(c/m)^2 * c^2 = (c^2+c^2/m^2) * a^2
Simplifying, we get a^2(m^2+1) = c^2
y = mx + c is equation for straight line. It touches the x axis at C(-c/m,0) and y axis at B(0,c).
For the line to touch the circle, it has to be a tangent as shown above (line BC).
From the figure, Area of Triangle ABC = 1/2 * AC * AB = 1/2 * BC * AD
BC^2 = AB^2 + AC^2 (Pythagoras Theorem)
So, squaring both sides, we have
(c/m)^2 * c^2 = (c^2+c^2/m^2) * a^2
Simplifying, we get a^2(m^2+1) = c^2
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