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Q.
Length of intercept made by the circle
$x^{2} + y^{2} - 16x + 4y - 36 = 0$ on $x$-axis is
Conic Sections
Solution:
As the circle meet $x$-axis at two points so put $y= 0$ in the equation of circle, we get $x^{2 }- 16x - 36 = 0$
Let two roots of the equation be $x_{1}$ and $x_{2}$.
$ \Rightarrow \, x_{1}+x_{2}=16$ and $x_{1}x_{2}=-36$
Now, $ x_{1}-x_{2}=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4x_{1}x_{2}}$
$=\sqrt{\left(16\right)^{2}+\left(4\times36\right)}=\sqrt{400}=20 =$ Intercept on $x$-axis