Question

A circle is dawn its centre on the line $x+y=2$ to touch the line $4x-3y+4=0$ and pass through the point $(0,1)$. Find the equation.

• A. $(x-1{)}^2+(y-1{)}^2=1$
• B. $(x-20{)}^2+(y+18{)}^2=841$
• C. $(x-21{)}^2+(y+19{)}^2=841$
• D. None

Answer 1

Answer: (A), (C)

$(x-1{)}^2+(y-1{)}^2=1$
$(x-21{)}^2+(y+19{)}^2=841$

Let the center of circle of circle be $(a,b)$ and radius$=r$

$\therefore x+y=2\Rightarrow a+b=2b=2-a$........(1)

Since circle touches the line: $4x-3y+4=0$, Hence ${\perp }^r$

distance of center from line will be equal to the radius of circle, hence

$\frac{|4a-3b+4|}{\sqrt{4^2+(-3{)}^2}}=r\frac{|4a-3(2-a)|}{5}$

Again the distance of centre $(a,b)$ from $pt(0,1)$

will be equal to the radius of circle.

$\therefore \sqrt{(a-0{)}^2+(b-1{)}^2}=r\Rightarrow \sqrt{a^2+(2-a-1{)}^2}=r$

$a^2+1+a^2-2a=r^2\Rightarrow 2a^2-2a+1=r^2$

From (ii)

$2a^2-2a+1={\left(\frac{|7a-2|}{5}\right)}^2$

$a^2-22a+21=0\Rightarrow a=1,21$

If $a=1\Rightarrow b=1,r=1$

If $a=21\Rightarrow b=-19,r=29$

$\therefore E{q}^n$ of circle is $2$

$(x-1{)}^2+(y-1)=1^2=1$

$(x-21{)}^2+(y+19{)}^2={29}^2=841$