Question

Equation of circle with centre (1,-3) and touches a line 2x-y-4=0 will be

• A. 5x${}^2$ + 5y${}^2$ - 10x + 30y + 40 = 0
• B. 5x${}^2$ + 5y${}^2$ - 10x + 30y + 49 = 0
• C. 5x${}^2$ + 5y${}^2$ - 10x + 30y - 40 = 0
• D. 5x${}^2$+ 5y${}^2$ - 10x + 30y - 49 = 0

Answer 1

Answer: (B)

5x${}^2$ + 5y${}^2$ - 10x + 30y + 49 = 0

Since the circle with center $(1,-3)$ touches the straight line: $2x-y-4=0$

Hence the length of perpendicular dropped from center $(1,-3)$ to the line: $2x-y-4=0$ will be equal to the radius of circle hence

radius$=\frac{|2\times 1-1\times -3-4|}{\sqrt{2^2+{(-1)}^2}}=\frac{1}{\sqrt{5}}$

Hence the equation of the circle is ${(x-1)}^2+{(y-(-3))}^2={\left(\frac{1}{\sqrt{5}}\right)}^2$

${(x-1)}^2+{(y+3)}^2={\left(\frac{1}{\sqrt{5}}\right)}^2$

$\Rightarrow x^2-2x+1+y^2+6y+9=\frac{1}{5}$

$\Rightarrow x^2+y^2-2x+6y+10=\frac{1}{5}$

$\Rightarrow 5x^2+5y^2-10x+30y+50-1=0$

$\therefore 5x^2+5y^2-10x+30y+49=0$ is the required equation of the circle.