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If $(4,7)$ and $(-2,-1)$ are ends of a diameter of a circle which intersects $X$-axis at $A$ and $B$, then $A B$ is equal to
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Verified Answer
The correct answer is:
8
End points of diameter are $(4,7)$ and $(-2,-1)$.
Then, equation of circle is
$$
\begin{aligned}
& (x-4)(x+2)+(y-7)(y+1)=0 \\
& \Rightarrow \quad x^2+y^2-2 x-6 y-15=0
\end{aligned}
$$
General equation of circle is
$$
x^2+y^2+2 g x+2 f y+c=0
$$
Compare Eqs. (i) and (ii), we obtain
$$
g=-1, f=-3, c=-15
$$
Now, $X$-intercept is given as,
$$
\begin{aligned}
A B & =2 \sqrt{g^2-c}=2 \sqrt{1-(-15)}=2 \sqrt{16}=8 \\
\therefore \quad A B & =8
\end{aligned}
$$
Then, equation of circle is
$$
\begin{aligned}
& (x-4)(x+2)+(y-7)(y+1)=0 \\
& \Rightarrow \quad x^2+y^2-2 x-6 y-15=0
\end{aligned}
$$
General equation of circle is
$$
x^2+y^2+2 g x+2 f y+c=0
$$
Compare Eqs. (i) and (ii), we obtain
$$
g=-1, f=-3, c=-15
$$
Now, $X$-intercept is given as,
$$
\begin{aligned}
A B & =2 \sqrt{g^2-c}=2 \sqrt{1-(-15)}=2 \sqrt{16}=8 \\
\therefore \quad A B & =8
\end{aligned}
$$
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