Question

Equation of circle which pass through the point $(1,-2)$ and $(3,-4)$ and touch the $x-axis$ is

• A. $x^2+y^2+6x+2y+9=0$
• B. $x^2+y^2+10x+20y+9=0$
• C. $x^2+y^2+61x+2y+90=0$
• D. $none$

Answer 1

The correct option is D $none$

Let $(h,k)$ be the centre of the circle

equation of circle : $(x-h{)}^2+(y-k{)}^2=r^2---\left(1\right)$

where $r=$radius of ncircle

$\therefore$ the circle touches $x-$axis

$\therefore$ It should pass through $(h,o)$ when it touches the $x-$axis and radius $r=k$

$\therefore$ from $\left(1\right),(x-h{)}^2+(y-k{)}^2=k^2----\left(2\right)$

$\because$ Circle pass through $(1,2)$ and $(3,-4)$

$\therefore (1-h{)}^2+(-2-4{)}^2=4^2$

or, $1-2h+h^2+4+4k+k^2=4^2$

or, h^2-2h+4h+5=0---(3)$

and $(3-h{)}^2+(-4-k{)}^2=4^2$

or, $9-6h+h^2+16+84+4^2=k^2$

lor, $h^2--6h+84+9=0----\left(4\right)$

Multiplying $\left(3\right)$ by $\left(2\right)$ and solving by $\left(3\right)-\left(4\right)$

$2h^2-4h+8k+10-h^2-6h-84-9=0$

or, $h^2+2h+1=0$

or, $(h+1{)}^2=0\Rightarrow h=1$

from $\left(4\right),(1{)}^2-6(1)+84+9=0\Rightarrow 1-6+9+8k=0$

$\Rightarrow 84=-4$

$\therefore 4=-1/2$

$\therefore$From $\left(2\right)$

$(x-1{)}^2+(y+1/2{)}^2=\frac{1}{4}$

or, $x^2-2x+1+y^2+y+\frac{1}{4}=\frac{1}{4}$

or, $x^2+y^2-2x+y=0.\Rightarrow$Required equation of circle.