Question

If the line $3x+4y=24$ and $4x+3y=24$ intersects the coordinates axes at $A,B,C$ and $D$, then the equation of the circle passing through these $4$ points is

• A. $x^2+y^2-12x-12y+48=0$
• B. $x^2+y^2-14x-14y+48=0$
• C. $x^2+y^2-12x-12y+46=0$
• D. $x^2+y^2-14x-14y+46=0$

Answer 1

Answer: (B)

$x^2+y^2-14x-14y+48=0$

writing in intercept form, we get
$\frac{x}{8}+\frac{y}{6}=1$ and
$\frac{x}{6}+\frac{y}{8}=1$
Hence the point are $(0,8),(0,6)$ and $(6,0)(8,0)$
Hence
The circle cuts the x axis at $(6,0)(8,0)$
Hence x-coordinate of the center will be
$=6+\frac{8-6}{2}$
$=7$
The circle cuts the x axis at $(0,6)(0,8)$
Hence y-coordinate of the center will be
$=6+\frac{8-6}{2}$
$=7$
The center of the circle will lie at
$(7,7)=C$
Now let the point $(6,0)$ be A.
Hence
R=radius
$CA$$=\sqrt{50}$
Hence, the equation of the circle will be
$(x-7{)}^2+(y-7{)}^2=50$
$x^2+y^2-14x-14y+98=50$
Or
$x^2+y^2-14x-14y+48=0$