Question

If the line $y=x+3$ meets the circle $x^2+y^2=a^2$ at $A$ and $B$, then the equation of the circle having $AB$ as a diameter will be ?

• A. $x^2+y^2+3x-3y-a^2+9=0$
• B. $x^2+y^2+3x+3y-a^2+9=0$
• C. $x^2+y^2-3x+3y-a^2+9=0$
• D. None of these

Answer 1

Answer: (A)

$x^2+y^2+3x-3y-a^2+9=0$

$[y=x+3]$

$[and{x}^2+y^2=a^2]are[x^2+(x+3{)}^2=a^2]$

$[2x^2+6x+9-a^2=0]$

$[x=(-6+-sqrt(36-8(9-a^2\left)\right)\left)/4\right]$

$[=(-3+-sqrt(9-2(9-a^2\left)\right))/2=(-3+-sqrt(2a^2-9)\left)/2\right]$

$=[y=3+(-3+-sqrt(2a^2-9))/2=(3+-sqrt(2a^2-9)\left)/2\right]$

$\left[A\right((-3+sqrt(2a^2-9\left)\right)/2,(3+sqrt(2a^2-9\left)\right)/2\left)\right]$

$\left[B\right((-3-sqrt(2a^2-9\left)\right)/2,(3-sqrt(2a^2-9\left)\right)/2\left)\right]$

$\left[\right(x+(3-sqrt(2a^2-9\left)\right)/2\left)\right(x+(3+sqrt(2a^2-9\left)\right)/2)+(y-(3+sqrt(2a^2-9\left)\right)/2\left)\right(y-(3-sqrt(2a^2-9\left)\right)/2)=0]$

$\left[\right(x+3/2{)}^2-(2a^2-9)/4+(y-3/2{)}^2-(2a^2-9)/4=0]$

$[x^2+3x+9/4+y^2-3y+9/4-a^2+18/4=0]$

$[x^2+y^2+3x-3y-a^2+9=0]$