# If 3x+2y=3 and 2x+5y=1 are conjugate lines with respect to the circle x^2+y^2=r^2 then r^2= A) 3/16 B) 16/3 C) 4/√3 D) √3/4

## Question

If $3x+2y=3$ and $2x+5y=1$ are conjugate lines with respect to the circle $x^{2}+y^{2}=r^{2}$ then $r^{2}=$

A) $\frac{3}{16}$ B) $\frac{16}{3}$ C) $\frac{4}{\sqrt{3}}$ D) $\frac{\sqrt{3}}{4}$

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## Answer

To find the value of ( r^2 ) given that the lines (3x + 2y = 3) and (2x + 5y = 1) are conjugate with respect to the circle ( x^2 + y^2 = r^2 ), we can use the condition for conjugate lines with respect to a circle.

### Condition for Conjugate Lines

For the lines ( l_1 x + m_1 y + n_1 = 0 ) and ( l_2 x + m_2 y + n_2 = 0 ) to be conjugate with respect to the circle ( x^2 + y^2 + 2gx + 2fy + c = 0 ), the condition is:

[ r^2 (l_1 l_2 + m_1 m_2) = (l_1 g + m_1 f - n_1)(l_2 g + m_2 f - n_2) ]

Given data:

- Circle equation: ( x^2 + y^2 = r^2 ) (so ( g = 0 ), ( f = 0 ), and ( c = r^2 ))
- Lines: ( 3x + 2y = 3 ) and ( 2x + 5y = 1 )

### Identifying the Coefficients

For the first line (( 3x + 2y = 3 )):

- ( l_1 = 3 )
- ( m_1 = 2 )
- ( n_1 = -3 ) (Note the sign change as the equation format is ( l_1 x + m_1 y + n_1 = 0 ))

For the second line (( 2x + 5y = 1 )):

- ( l_2 = 2 )
- ( m_2 = 5 )
- ( n_2 = -1 ) (Again, note the sign change)

### Substituting into the Equation

Using the given condition:

[ r^2 (l_1 l_2 + m_1 m_2) = (l_1 g + m_1 f - n_1)(l_2 g + m_2 f - n_2) ]

Substitute the given values here:

- ( g = 0 )
- ( f = 0 )

So the right-hand side becomes:

[ (3 \cdot 0 + 2 \cdot 0 - (-3))(2 \cdot 0 + 5 \cdot 0 - (-1)) ] [ = (-(-3))(-(-1)) ] [ = 3 \cdot 1 ] [ = 3 ]

The left-hand side is:

[ r^2 (3 \cdot 2 + 2 \cdot 5) ] [ = r^2 (6 + 10) ] [ = r^2 \cdot 16 ]

### Equating Both Sides

Set these equal to each other:

[ r^2 \cdot 16 = 3 ]

Solving for ( r^2 ):

[ r^2 = \frac{3}{16} ]

Therefore, the correct option is:

[ \boxed{\frac{3}{16}} ]

So, the value of ( r^2 ) is **( \frac{3}{16} )**, which corresponds to option **A**.

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