Question

A circle of radius $a$ with both coordinates of its centre positive, touches the $x-axis$ and also the line $3y=4x$ . Then its equation is :

& \left( A \right){{x}^{2}}+{{y}^{2}}-2ax-2ay+{{a}^{2}}=0 \\\ & \left( B \right){{x}^{2}}+{{y}^{2}}-6ax-4ay+12{{a}^{2}}=0 \\\ & \left( C \right){{x}^{2}}+{{y}^{2}}-4ax-2ay+4{{a}^{2}}=0 \\\ & \left( D \right)\text{none of the foregoing equations} \\\ \end{aligned}$$

Answer 1

We will be considering the centre coordinates as $\left( a,a \right)$ as the circle has a positive centre. We are told that the circle touches the $x-axis$ as well as the line $3y=4x$. So we can understand that the line becomes the tangent to the circle. Next we have to find out the distance between the centre and the line given and equate it to $a$ as $a$ is the radius of the circle. And upon substituting the values we get the required equation of the circle.

Complete step-by-step solution:
Now let us learn about finding circle equations. We know that the general equation for a circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ where $\left( h,k \right)$ is centre of the circle and $r$ is the radius of the circle. If the centre of the circle is the origin itself, then the equation of the circle is ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$.
Now let us find out the required circle equation.
We plot the rough figure below such that the circle touches line $3y=4x$ and also the positive $x-axis$ . Also, the radius is $a$. So, we have to consider the y-coordinate as a. As we know that the centre is positive, we will be considering it as $\left( \alpha,a \right)$.

Now the distance from the centre to the line would be equal to radius i.e. $a$
$\Rightarrow r=a$
The line equation can be written as $4x-3y=0$.
We know that the perpendicular distance from the center of a circle, say $(x’,y’)$ to the tangent $ax+by+c=0$ is radius.
$\text{Radius}=\dfrac{|ax’+by’+c|}{\sqrt{a^2+b^2}}$
So, here the line is tangential to the circle. So, using the formula for perpendicular distance and equating it to radius, we get,

& \Rightarrow \dfrac{4\alpha -3a}{\sqrt{{{4}^{2}}+{{\left( -3 \right)}^{2}}}}=a \\\ & \Rightarrow \dfrac{4\alpha -3a}{\sqrt{16+9}}=a \\\ & \Rightarrow \dfrac{4\alpha -3a}{\sqrt{25}}=a \\\ & \Rightarrow 4\alpha -3a=5a \\\ & \Rightarrow 4\alpha =8a \\\ & \Rightarrow \alpha =2a \\\ \end{aligned}$$ Now we can express the centre as $$\left( 2a,a \right)$$. The general form of the equation of the circle when centre is given is as $$\left( a,b \right)$$ is $${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$$. Now similarly applying this formula to our problem, we get the equation as $$\begin{aligned} & \Rightarrow {{\left( x-2a \right)}^{2}}+{{\left( y-a \right)}^{2}}={{a}^{2}} \\\ & \Rightarrow {{x}^{2}}+4{{a}^{2}}-4ax+{{y}^{2}}+{{a}^{2}}-2ay={{a}^{2}} \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-4ax-2ay+4{{a}^{2}}=0 \\\ \end{aligned}$$ **$$\therefore $$ The equation of the circle of radius $$a$$ with both coordinates of its centre positive, touches the $$x-axis$$ and also the line$$3y=4x$$is $${{x}^{2}}+{{y}^{2}}-4ax-2ay+4{{a}^{2}}=0$$.** **Note:** We must note that a line is tangent only when it is perpendicular to the radius of the circle. We can say that if two tangents are drawn from the same external point to the circle, then the tangents are congruent. We can find the common tangents to the circle with the formula $$S{{S}_{1}}=T$$where $$S$$ is the equation of one circle.