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  • Consider a family of circles passing through two fixed points A(3,7) and B(6,5) .show that the chords in which the circle <img alt="\mathrm{x^2+y^2-4 x-6 y-3=0}" src="https://entrancecorner.on

Consider a family of circles passing through two fixed points A(3,7) and B(6,5) .show that the chords in which the circle \mathrm{x^2+y^2-4 x-6 y-3=0} cuts the members of the family are cocurrent at a point .,The coorinate of this point is 

Option: 1

\mathrm{\left(2, \frac{23}{3}\right)}


Option: 2

\mathrm{\left(1, \frac{23}{3}\right)}


Option: 3

\mathrm{\left(2, \frac{3}{23}\right)}


Option: 4

\mathrm{\left(3, \frac{23}{3}\right)}


Answers (1)

best_answer

: Let the circle be \mathrm{x^2+y^2+2 g x+2 f y+c=0}

It passes through A(3, 7) and B(6, 5)

\mathrm{\begin{aligned} & \therefore 3^2+7^2+6 g+14 f+c=0 \\ & \Rightarrow 58+6 g+14 f+c=0 \\ & \text { and } 6^2+5^2+12 g+10 f+c=0 \\ & \Rightarrow 61+12 g+10 f+c=0 \end{aligned}}-----------(1), (2)

Equation (2) and (1) gives

\mathrm{\begin{aligned} & 3+6 g-4 f=0 \\ & \therefore 6 g=4 f-3 \text { or } 2 g=\frac{4 f-3}{3} \end{aligned}}

Hence from (1)

\mathrm{\begin{aligned} & c=-[58+14 f+6 g]=-\left[58+14 f+3 \times \frac{4 f-3}{3}\right] \\ & \therefore c=-[55+18 f] \end{aligned}}

Hence the family of circles is

\mathrm{\begin{aligned} & S_1=x^2+y^2+\frac{4 f-3}{3} x+2 f y-(55+18 f)=0 \\ & S_2=x^2+y^2-4 x-6 y-3=0 \end{aligned}}

Common chord is given by

\mathrm{\begin{aligned} & S_1-S_2=0 \\ & \Rightarrow\left(\frac{4 f-3}{3}+4\right) x+(2 f+6) y-(55+18 f-3)=0 \\ & \Rightarrow(3 x+6 y-52)+f\left(\frac{4}{3} x+2 y-18\right)=0 \end{aligned}}

Above is of the form P + λ Q = 0, which passes through
the intersection of P = 0 and Q = 0

Solving 3x + 6y – 52 = 0 and 4x + 6y – 54 = 0

we get x = 2 and \mathrm{y=\frac{23}{3}}

∴ Required point is \mathrm{\left(2, \frac{23}{3}\right)}

Posted by

HARSH KANKARIA

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