Let a given line L₁ intersects the x...

MathematicsStandard and General Equation of a CircleJEE Advanced 1987Moderate

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

Visualized Solution (English)

Defining Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{1} and Intercepts

Let line $L_{1}$ be $\frac{x}{a} + \frac{y}{b} = 1$
The intercepts are $P (a, 0)$ and $Q (0, b)$
Slope of $L_{1}$ is $m_{1} = - \frac{b}{a}$

Defining Perpendicular Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{2}

$L_{2} ⊥ L_{1} ⟹$ Slope of $L_{2}$ is $m_{2} = \frac{a}{b}$
Let $L_{2}$ be $y = \frac{a}{b} x + c$
Intercepts of $L_{2}$ : $R (- \frac{b c}{a}, 0)$ and $S (0, c)$

Equations of Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P S and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Equation of $P S$ (through $P (a, 0), S (0, c)$ ): $\frac{x}{a} + \frac{y}{c} = 1 ⟹ c x + a y = a c$
Equation of $QR$ (through $Q (0, b), R (- \frac{b c}{a}, 0)$ ): $\frac{x}{- b c / a} + \frac{y}{b} = 1 ⟹ - a x + cy = b c$

Eliminating the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

From $P S$ : $c (x - a) = - a y ⟹ c = \frac{a y}{a - x}$
Substitute $c$ into $QR$ : $- a x + \frac{a y ^{2}}{a - x} = \frac{ab y}{a - x}$

Simplifying to the Final Locus

Multiply by $(a - x)$ : $- a x (a - x) + a y^{2} = ab y$
Divide by $a$ : $- a x + x^{2} + y^{2} = b y$
Rearranging: $x^{2} + y^{2} - a x - b y = 0$

Conclusion and Origin Check

The locus is $x^{2} + y^{2} - a x - b y = 0$
At $(0, 0)$ : $0^{2} + 0^{2} - a (0) - b (0) = 0$
Conclusion: The locus is a circle passing through the origin.

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(B)

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(C)

{(x, y) : x^{2} = y} \cup {(x, y) : y \leq 0}

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Visualized Solution (English)

Defining Line $L_{1}$ and Intercepts

Defining Perpendicular Line $L_{2}$

Equations of Lines $P S$ and $QR$

Eliminating the Parameter $c$

Simplifying to the Final Locus

Conclusion and Origin Check

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Visualized Solution (English)

Defining Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and Intercepts

Defining Perpendicular Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Equations of Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PS and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } QR

Eliminating the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Simplifying to the Final Locus

Conclusion and Origin Check

Conceptually Similar Problems

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Through a fixed point (h,k) secants are drawn to the circle x2+y2=r2. Show that the locus of the mid-points of the secants intercepted by the circle is x2+y2=hx+ky.

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Defining Line $L_{1}$ and Intercepts

Defining Perpendicular Line $L_{2}$

Equations of Lines $P S$ and $QR$

Eliminating the Parameter $c$

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = k^{2}$ orthogonally, then the equation of the locus of its centre is

Through a fixed point $(h, k)$ secants are drawn to the circle $x^{2} + y^{2} = r^{2}$ . Show that the locus of the mid-points of the secants intercepted by the circle is $x^{2} + y^{2} = h x + k y$ .

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = p^{2}$ orthogonally, then the equation of the locus of its centre is

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the locus of its centre is

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = k^{2}$ orthogonally, then the equation of the locus of its centre is

Through a fixed point $(h, k)$ secants are drawn to the circle $x^{2} + y^{2} = r^{2}$ . Show that the locus of the mid-points of the secants intercepted by the circle is $x^{2} + y^{2} = h x + k y$ .

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = p^{2}$ orthogonally, then the equation of the locus of its centre is

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the locus of its centre is

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .