Question

An equation of the circle in which the chord joining the points $(1,2)$ and $(2,-1)$ subtends an angle of $\frac{\pi }{4}$ at any point on the circumference is

• A. $x^2+y^2-15=0$
• B. $x^2+y^2-6x-2y+5=0$
• C. $x^2+y^2+6x+2y-15=0$
• D. $x^2+y^2-2x-4y+4=0$

Answer 1

The correct option is B $x^2+y^2-6x-2y+5=0$

Let $P(x,y)$ be any point on the circle passing through $A(1,2)$ and $B(2,-1)$

The slope of $AP=\frac{y-2}{x-1}$ and slope of $BP=\frac{y+1}{x-2}$

Since they include an angle of $\frac{\pi }{4}$, so

$\frac{\frac{y-2}{x-1}-\frac{y+1}{x-2}}{1+\frac{y-2}{x-1}\times \frac{y+1}{x-2}}=\pm \tan \frac{\pi }{4}=\pm 1\Rightarrow (y+1)(x-1)-(x-2)(y-2)=\pm [(x-1)(x-2)+(y-2)(y+1)]\Rightarrow 3x+y-5=\pm [x^2+y^2-3x-y]$

$\Rightarrow x^2+y^2-5=0$ and $x^2+y^2-6x-2y+5=0$