Question

A circle has radius $3$ units and its centre lies on the line $y=x-1$. if it passes through $(7,3)$, its equation

• A. $x^2+y^2-8x-14y=0$
• B. $x^2+y^2-8x-6y-6=0$
• C. $x^2+y^2-14x-12y-76=0$
• D. $x^2+y^2+14x+12x-70=0$

Answer 1

The correct option is A $x^2+y^2-8x-14y=0$

${(x-h)}^2+{(y-k)}^2=r^2$

Here $(h,k)\to$ centre

Equation of line through centre

$k=h-1----\left(1\right)$

& $(x,y)$ point on circle

So, given $(x,y)=(7,3)$

So, ${(7-h)}^2+{(3-h+1)}^2=3^2$

Using eqn $\left(1\right)$

We get,

$28+h^2-11h=0$

$h=7$ oR $4$

$k=6$ oR $3$

Equation of circle corresponding $(7,6)$

$x^2+y^2-14x-12y+76=0$

Equation of circle corresponding $(4,3)$

$x^2+y^2-8x-6y+16=0$