The pole of 3x + 5y + 17 = 0 with respect to the circle x^2 + y^2 + 4x + 6y + 9 = 0 is a) (1,2) b) (1,2) c) (1,2) d) (2,1)
Question
The pole of $3x + 5y + 17 = 0$ with respect to the circle $x^2 + y^2 + 4x + 6y + 9 = 0$ is
a) $(1,2)$
b) $(1,2)$
c) $(1,2)$
d) $(2,1)$
✨ Free AI Tools for You
Powered by Chatterbot AI
Answer
To determine the pole of the line $3x + 5y + 17 = 0$ with respect to the circle $x^2 + y^2 + 4x + 6y + 9 = 0$, follow these steps:

Identify the given circle and line:
 Circle: $x^2 + y^2 + 4x + 6y + 9 = 0$
 Line: $3x + 5y + 17 = 0$

Let ( P(\alpha, \beta) ) be the pole of the line with respect to the circle.

Use the formula for the equation of the pole of point ( (\alpha, \beta) ) with respect to the circle: $$ x\alpha + y\beta + 2(x + \alpha) + 3(y + \beta) + 9 = 0 $$

Simplify the equation: $$ x\alpha + y\beta + 2x + 2\alpha + 3y + 3\beta + 9 = 0 $$ Combine like terms: $$ (2 + \alpha)x + (3 + \beta)y + (2\alpha + 3\beta + 9) = 0 $$

This should be identical to the given line: $$ 3x + 5y + 17 = 0 $$

Match the coefficients:
 For (x): ( 3 = 2 + \alpha \implies \alpha = 1 )
 For (y): ( 5 = 3 + \beta \implies \beta = 2 )
 The constant term should match naturally if above values are correct but you can verify if needed: $$ 2\alpha + 3\beta + 9 = 2(1) + 3(2) + 9 = 2 + 6 + 9 = 17 $$

Thus, the pole is ( P(1, 2) ).

Correct answer:
 (b) ((1, 2))
Therefore, the pole of the line $3x + 5y + 17 = 0$ with respect to the circle $x^2 + y^2 + 4x + 6y + 9 = 0$ is ( \boxed{(1, 2)} ).
India's 1^{st} AI Doubt Solver for CBSE, JEE, and NEET
Ask a Question for Freeand then it's just ₹212 a month