Question

The circle $C$ passing through the origin, has the line $3x+4y=0$ as its tangent at the origin. If the image of the centre of $C$ w.r.t the line $\frac{x}{\frac{25}{3}}+\frac{y}{\frac{25}{4}}=1$ lies on the line $3x+4y=0,$ then the equation of $C$ is

• A. $(x-6{)}^2+(y-8{)}^2=100$
• B. $(x-6{)}^2+(y-0{)}^2=100$
• C. $x^2+y^2=100$
• D. $(x+6{)}^2+(y+8{)}^2=100$

Answer 1

The correct option is A $(x-6{)}^2+(y-8{)}^2=100$

Equation of the line which contains the origin and centre of the circle is perpendicular to line $3x+4y=0$
$\therefore \text{The equation of the line is}4x-3y=0$
$\because 4x-3y=0$ is also perpendicular to $\frac{x}{\frac{25}{3}}+\frac{y}{\frac{25}{4}}=1$
$\therefore$ Image of the the centre of the circle w.r.t line $\frac{x}{\frac{25}{3}}+\frac{y}{\frac{25}{4}}=1$ is origin.
Distance of the line from the origin is $5$
$\because \text{centre lies on the line}4x-3y=0\text{and at a distance of}5\text{from}\frac{x}{\frac{25}{3}}+\frac{y}{\frac{25}{4}}=1$
So, centre of the circle is $(6,8)$
So, the equation of the circle is $(x-6{)}^2+(y-8{)}^2=100$