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Circles

**Correct Marks**
2

**Incorrectly Marks**
-0.5

Q 1. If the lines {tex} 2x + 3y + 1 = 0 {/tex} and {tex} 3x - y - 4 = 0 {/tex} lies along diameter of a circle of circumference {tex} 10 \pi {/tex}, then the equation of the circle is:

{tex} x^{2} + y^{2} - 2x + 2y - 23 = 0 {/tex}

{tex} x^{2} + y^{2} - 2x - 2y - 23 = 0 {/tex}

{tex} x^{2} + y^{2} + 2x + 2y - 23 = 0 {/tex}

{tex} x^{2} + y^{2} + 2x - 2y - 23 = 0 {/tex}

The lines {tex} 2x+3y+1=0\ and\ 3x-y-4=0 {/tex} are diameter of circle

On solving these equations, we get

{tex} x=1\ and\ y=-1 {/tex}

Therefore the centre or circle = {tex} = \left(1-1\right) {/tex}) and

Circumference = 10{tex} \pi {/tex}

=> {tex} 2 \pi r = 10 \pi => r = 5 {/tex}

Equation of circle

{tex} \left(x-x_{1}\right)^{2}+ \left(y-y_{1}\right)^{2}=r^{2} {/tex}

=> {tex} \left(x-1\right)^{2}+ \left(y+1\right)^{2}=5^{2} {/tex}

=> {tex} x^{2}+1-2x+y^{2}+2y+1=25 {/tex}

=> {tex} x^{2}+y^{2}-2x+2y-23=0 {/tex}

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