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Standard XII

Mathematics

Equation of Circle with Origin as Center

Let a given l...

Question

Let a given line $L_{1}$ intersect the $x$ and $y -$ axis at $P$ and $Q$ , respectively. Let another line $L_{2}$ , perpendicular to $L_{1}$ , cut the $x -$ axis and $y -$ axis at $R$ and $S$ , respectively. Show that the locus of the point of intersection of the lines $P S$ and $Q R$ , is a circle passing through the origin ?

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Solution

$(R, S) = (k m, 0), (0, k)$ , $(Q, R) = (k m, 0), (0, k)$

$y = \frac{- c x}{k m} + c$ QR is $y = \frac{- x}{k m} + c$

From PS and QR equations,

$k = \frac{c y}{(m x + c)} = \frac{c x}{(c m - m y)}$

$c y (c m - m y) = c x (m x + c)$

$y (c m - m y) = x (m x + c)$

$m x^{2} + m y^{2} + c x - m c y = 0$

$x^{2} + y^{2} + \frac{c}{m} x - c y = 0$ , is the equation of circle passing through the origin.

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Let a given line L1 intersect the x and y−axis at P and Q, respectively. Let another line L2, perpendicular to L1, cut the x−axis and y−axis at R and S, respectively. Show that the locus of the point of intersection of the lines PS and QR, is a circle passing through the origin ?

(R,S)=(km,0),(0,k) , (Q,R)=(km,0),(0,k)y=−cxkm+c QR is y=−xkm+cFrom PS and QR equations,k=cy(mx+c)=cx(cm–my)cy(cm–my)=cx(mx+c)y(cm–my)=x(mx+c)mx2+my2+cx–mcy=0x2+y2+cmx–cy=0, is the equation of circle passing through the origin.