Question

The circle which passes through the points $(1,-2)$ and $(4,-3)$ and whose centre lies on the line $3x+4y=7$ is

• A. $15(x^2+y^2)-94x+18y+55=0$
• B. $15(x^2+y^2)+94x+18y+55=0$
• C. $15(x^2+y^2)-94x-18y+55=0$
• D. $15(x^2+y^2)-94x-18y-55=0$

Answer 1

The correct option is A $15(x^2+y^2)-94x+18y+55=0$

The line through $(1,-2)$ and $(4,-3)$ is given by
$\frac{y+2}{x-1}=-\frac{1}{3}$
or $x+3y+5=0$
Using the above information of
Circle through the two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)+\lambda L=0,\lambda \in R$ we get
The circle is $(x-1)(x-4)+(y+2)(y+3)+\lambda (x+3y+5)=0$
or $x^2+y^2+(\lambda -5)x+(5+3\lambda )y+5(\lambda +2)=0$
The centre is $(\frac{5-\lambda }{2},\frac{-5-3\lambda }{2})$ lies on the line $3x+4y=7$
$\therefore \frac{3}{2}(5-\lambda )-2(5+3\lambda )=7$
$\Rightarrow \lambda =-\frac{19}{15}$
$\therefore$ The circle is $x^2+y^2-5x+5y+10-\frac{19}{15}(x+3y+5)=0$
or $15x^2+15y^2-75x+75y+150-19x-57y-95=0$
or $15(x^2+y^2)-94x+18y+55=0$