If the circle x^2 + y^2 = a^2 cuts a chord of | vedclass

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(B) The perpendicular distance $d$ from the center $(0,0)$ to the line $mx - y + c = 0$ is given by:$d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}}

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$(1 + m^2)(a^2 - b^2) = c^2$

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(B) The perpendicular distance $d$ from the center $(0,0)$ to the line $mx - y + c = 0$ is given by:$d = \frac{|m(0) - (0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{1 + m^2}}$In the right-angled triangle formed by the radius $a$,the half-chord $b$,and the perpendicular distance $d$,we have by Pythagoras theorem:$a^2 = b^2 + d^2$$d^2 = a^2 - b^2$Substituting the value of $d$:$\left(\frac{|c|}{\sqrt{1 + m^2}}\right)^2 = a^2 - b^2$$\frac{c^2}{1 + m^2} = a^2 - b^2$$c^2 = (1 + m^2)(a^2 - b^2)$

If the circle $x^2 + y^2 = a^2$ cuts a chord of length $2b$ on the line $y = mx + c$,then:

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