Question

Equation of the circle which touches $3x+4y=7$ and passes through $(1,-2)$ and $(4,-3)$ is

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

Answer 1

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

$\begin{array}{c}Generalequofcircle:\underset{–}{x^2+y^2+2gx+2fy+c=0}\\ Now,theequispassesthrough(1,-2)\\ \Rightarrow 1+4+2g-4f+c=0\\ \therefore 5+2g-4f+c=0------\left(i\right)\\ Again,passesthrough(4,-3)\\ \Rightarrow 16+9+8g-6f+c=0\\ \therefore 25+8g-6f+c=0-------\left(ii\right)\\ Since,thecirclecenteris:(-g,-f)substituteinline3x+4y=7\\ \Rightarrow 3(-g)+4(-f)=7--------\left(iii\right)\\ \Rightarrow -3g-4f=7\\ \therefore g=\frac{4f-7}{3}\\ ifsubstitutevalueof{}^{\prime }{g}^{\prime }intoequ\left(i\right)\&\left(ii\right),wegetequintermoff\&c.\\ thenwefind,\\ g=\frac{-47}{15},f=\frac{3}{5},c=\frac{11}{3}\\ \sin ce,substitutethesevaluein:\\ x^2+y^2+2gx+2fy+c=0\\ \Rightarrow x^2+y^2+2\times \left(\frac{-47}{15}\right)x+2\left(\frac{3}{5}\right)y+\frac{11}{3}=0\\ \Rightarrow x^2+y^2-\frac{94}{15}x+\frac{6}{5}y+\frac{11}{3}=0\\ \therefore 15x^2+15y^2-94x+18y+55=0\\ So,thatthecorrectoptionisB.\end{array}$