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A line moves such that the portion of it intercepted between the coordinate axes is of constant length \(a\), then the locus of the mid point of that line segment is
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Verified Answer
The correct answer is:
\(x^2+y^2=\frac{a^2}{4}\)
Let \(A=(p, 0)\)
\(B=(0, q)\)
Let \( P=(h, k)\) be the mid-point of \(\overline{A B}\).
\(\begin{array}{rlrl}
\text {Given, } \overline{A B} & =a \\
(h, k) & =\text { mid-point of } \overline{A B} \\
(h, k) & =\left(\frac{p}{2}, \frac{q}{2}\right) \\
\therefore \quad & p & =2 h, q=2 k \\
\therefore \quad & A & =(p, 0)=(2 h, 0) \\
& B & =(0, q)=(0,2 k)
\end{array}\)
Since, length of \(A B=a\)
\(\begin{aligned}
\sqrt{(2 h)^2+(2 k)^2} & =a \\
\sqrt{4 h^2+4 k^2} & =a \\
4 h^2+4 k^2 & =a^2 \\
h^2+k^2 & =\frac{a^2}{4}
\end{aligned}\)
\(\therefore\) Required locus is \(x^2+y^2=\frac{a^2}{4}\)
\(\therefore\) Hence, solution is (c).
\(B=(0, q)\)
Let \( P=(h, k)\) be the mid-point of \(\overline{A B}\).
\(\begin{array}{rlrl}
\text {Given, } \overline{A B} & =a \\
(h, k) & =\text { mid-point of } \overline{A B} \\
(h, k) & =\left(\frac{p}{2}, \frac{q}{2}\right) \\
\therefore \quad & p & =2 h, q=2 k \\
\therefore \quad & A & =(p, 0)=(2 h, 0) \\
& B & =(0, q)=(0,2 k)
\end{array}\)
Since, length of \(A B=a\)
\(\begin{aligned}
\sqrt{(2 h)^2+(2 k)^2} & =a \\
\sqrt{4 h^2+4 k^2} & =a \\
4 h^2+4 k^2 & =a^2 \\
h^2+k^2 & =\frac{a^2}{4}
\end{aligned}\)
\(\therefore\) Required locus is \(x^2+y^2=\frac{a^2}{4}\)
\(\therefore\) Hence, solution is (c).
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