Question:
The pole of the line \(\dfrac{x}{a}+\dfrac{y}{b}=1\) with respect to the circle \(x^2+y^2=c^2\) is
The pole of the line \(\dfrac{x}{a}+\dfrac{y}{b}=1\) with respect to the circle \(x^2+y^2=c^2\) is
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For the circle \(x^2+y^2=c^2\), the polar of \((x_1,y_1)\) is \(xx_1+yy_1=c^2\). Compare the given line with this form to find the pole.
Updated On: Jun 15, 2026
- \(\left(\dfrac{c^2}{a},\dfrac{c^2}{b}\right)\)
- \(\left(\dfrac{c^2}{b},\dfrac{c^2}{a}\right)\)
- \(\left(\dfrac{c}{a},\dfrac{c}{b}\right)\)
- \(\left(\dfrac{c}{b},\dfrac{c}{a}\right)\)
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The Correct Option is A
Solution and Explanation
Step 1: Recall the polar form for a circle.
For the circle \[ x^2+y^2=c^2 \] the polar of a point \((x_1,y_1)\) is given by \[ xx_1+yy_1=c^2 \]
Step 2: Compare the given line with the polar equation.
The given line is \[ \frac{x}{a}+\frac{y}{b}=1 \] Multiplying by \(c^2\), we can write it as \[ x\cdot \frac{c^2}{a}+y\cdot \frac{c^2}{b}=c^2 \]
Step 3: Identify the pole.
Comparing \[ xx_1+yy_1=c^2 \] with \[ x\cdot \frac{c^2}{a}+y\cdot \frac{c^2}{b}=c^2, \] we get \[ x_1=\frac{c^2}{a},\qquad y_1=\frac{c^2}{b} \] Therefore, the pole of the given line is \[ \left(\frac{c^2}{a},\frac{c^2}{b}\right) \]
Step 4: Final conclusion.
Hence, the required pole is \[ \boxed{\left(\frac{c^2}{a},\frac{c^2}{b}\right)} \]
For the circle \[ x^2+y^2=c^2 \] the polar of a point \((x_1,y_1)\) is given by \[ xx_1+yy_1=c^2 \]
Step 2: Compare the given line with the polar equation.
The given line is \[ \frac{x}{a}+\frac{y}{b}=1 \] Multiplying by \(c^2\), we can write it as \[ x\cdot \frac{c^2}{a}+y\cdot \frac{c^2}{b}=c^2 \]
Step 3: Identify the pole.
Comparing \[ xx_1+yy_1=c^2 \] with \[ x\cdot \frac{c^2}{a}+y\cdot \frac{c^2}{b}=c^2, \] we get \[ x_1=\frac{c^2}{a},\qquad y_1=\frac{c^2}{b} \] Therefore, the pole of the given line is \[ \left(\frac{c^2}{a},\frac{c^2}{b}\right) \]
Step 4: Final conclusion.
Hence, the required pole is \[ \boxed{\left(\frac{c^2}{a},\frac{c^2}{b}\right)} \]
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