Question

Find the equation of a circle which has the portion the line $3x+4y=14$ intercepted by the lines $x-y=0$ and $11x-4y=0$ as a diameter.

Answer 1

We have,

Equation of intercepted line is

$3x+4y=14......\left(1\right)$

Equation of lines diameter

$x-y=0......\left(2\right)$

$11x-4y=0......\left(3\right)$

From equation (1) and (2) to, and we get,

$(x_1,y_1)=(7,7)$

From equation (1) and (3) to, and we get,

$(x_2,y_2)=(1,\frac{11}{4})$

So, equation of circle if given diameter points

Then,

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$\Rightarrow (x-7)(x-1)+(y-7)(y-\frac{11}{4})=0$

$\Rightarrow x^2-x-7x+7+y^2-\frac{11}{4}y-7y+\frac{77}{4}=0$

$\Rightarrow x^2+y^2-8x-\frac{39}{4}y+\frac{105}{4}=0$

$\Rightarrow 4x^2+4y^2-8x-39y+105=0$

Hence, this is the answer.