Question

If circle $S\left( x,y \right)=0$ touches the line $x+y-5=0$ at $\left( 2,3 \right)$ and $S\left( 1,2 \right)=-2$ the radius of the circle is

Answer 1

From the question given we have to find the radius of the circle of $S\left( x,y \right)=0$. To find the radius of circle, firstly we know the general equation for the intersection of line and circle is ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}+\lambda \left(\text{line equation} \right)=0$ where $\lambda$ is any constant after finding the value of $\lambda$. We will get a circle equation and then for the general circle equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ to find the radius the formula is $\text{radius}=\left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)$

Complete step-by-step solution:
From the given question the circle $S\left( x,y \right)=0$ touches the line $x+y-5=0$ at $\left( 2,3 \right)$
Now, after comparing with the general equation of intersection of line and circle. The circle equation can be written.
And from the question we also know that the $S\left( 1,2 \right)=-2$.
We also know that $\left( 2,3 \right)$, is a point on the circle
By this we can say that,
$\Rightarrow \left( h,k \right)=\left( 2,3 \right)$
As we know that the general equation of circle for the intersection of line and circle is
$\Rightarrow {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}+\lambda \left(\text{line equation} \right)=0$
In the question itself we know the equation of line that is
$\Rightarrow \text{line equation}=x+y-5=0$
Now by substituting in their respective positions in the circle equation we will get,
$\Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}+\lambda \left( x+y-5 \right)=0$
As we already know that $S\left( 1,2 \right)=-2$.
So, by using this we will get value of $\lambda$
By substituting the values, we will get,
$\Rightarrow {{\left( 1-2 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}+\lambda \left( 1+2-5 \right)=-2$
$\Rightarrow {{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+\lambda \left( -2 \right)=-2$
$\Rightarrow -2\lambda =-4$
$\Rightarrow \lambda =2$
Therefore, we got the value of $\lambda$is $2$
With this we got the circle equation that is
$\Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}+2\left( x+y-5 \right)=0$
By simplifying further
$\Rightarrow {{x}^{2}}+4-4x+{{y}^{2}}-6y+9+2x+2y-10=0$
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2x-4y+3=0$
Therefore, the circle equation is ${{x}^{2}}+{{y}^{2}}-2x-4y+3=0$
As we know that the radius of the general equation of a circle is
$\Rightarrow {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$
For the above equation the radius is
$\Rightarrow \text{radius} =\left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)$
By this the radius of the $S\left( x,y \right)=0$ is
By comparing the results are
$\Rightarrow g=-1$
$\Rightarrow f=-2$
$\Rightarrow c=3$
By substituting these values in the above formula, we will get radius
$\Rightarrow \text{radius}=\left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)$
$\Rightarrow r=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}-3}$
$\Rightarrow r=\sqrt{1+4-3}$
$\Rightarrow r=\sqrt{2}$
Therefore, the radius of the circle is $r=\sqrt{2}$
The figure for the question will be as follows.

Note: Students should know the concept of circle like, the general circle equation ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ and its radius and centre is $\text{centre}=\left( -g,-f \right)$, $\text{radius}=\left( \sqrt{{{g}^{2}}+{{f}^{2}}-c} \right)$. The general equation of a circle whose centre is origin is ${{x}^{2}}+{{y}^{2}}={{r}^{2}}$ where r is the radius of the circle.