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8 Questions around this concept.
Find the X-intercept and Y-intercept of line 3x-4y + 12 = 0.
Radius of the circle with centre $(3,-1)$ and cutting a chord of length 6 on the line $2 x-5 y+18=0$ is
Length of Intercept Cut-Off from a line
The length of the intercept cut off from the line $L: y=m x+c$ by the circle $x^2+y^2=a^2$ is
$
2 \times \sqrt{\left(\frac{a^2\left(1+m^2\right)-c^2}{\left(1+m^2\right)}\right)}
$
$
\begin{aligned}
& \mathrm{OM}=\left|\frac{\mathrm{c}}{\sqrt{1+\mathrm{m}^2}}\right| \\
& \text { In } \Delta \mathrm{OAM}, \quad \mathrm{AM}^2=\mathrm{AO}^2-\mathrm{OM}^2 \\
& =\mathrm{a}^2-\frac{\mathrm{c}^2}{1+\mathrm{m}^2} \\
& =\frac{a^2\left(m^2+1\right)-c^2}{\left(1+m^2\right)} \\
& \Rightarrow \quad \mathrm{AM}=\sqrt{\frac{\mathrm{a}^2\left(\mathrm{~m}^2+1\right)-\mathrm{c}^2}{\left(1+\mathrm{m}^2\right)}} \\
& \text { Length of intercept is }=\mathrm{AB} \\
& =2 \mathrm{AM} \\
& \mathrm{AB}=2 \times \sqrt{\frac{\mathrm{a}^2\left(\mathrm{~m}^2+1\right)-\mathrm{c}^2}{\left(1+\mathrm{m}^2\right)}}
\end{aligned}
$
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