
Find the equation of a circle that touches each of the axes and therefore the line 3x – 4y + 8 = 0 and lies within the third quadrant.
Answer
161.4k+ views
Hint: We will apply the formula of perpendicular distance between the points and the line. Hence, we will get the coordinates of the circle.
Formula Used: \[\begin{array}{*{20}{c}}
d& = &{\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}}
\end{array}\]
Complete step by step solution: The circle lies in the third quadrant. So, the coordinates of the center of the circle will be negative. If the circle touches both the axis, then the distance of the center point from the X and Y axis will be equal.
Let us assume that the coordinate of the circle is O(-m, -m).
According to the given problem, we will draw a figure,

Figure 1
In this problem, we have given an equation 3x – 4y + 8 = 0. Therefore, we will apply the formula of perpendicular distance between the points and the line.
Now,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}}
\end{array}\]
Therefore, we will put the values in the above equation
\[\begin{array}{*{20}{c}}
{ \Rightarrow OD}& = &{\dfrac{{3( - m) - 4( - m) + 8}}{{\sqrt {9 + 16} }}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow OD}& = &{\dfrac{{3( - m) - 4( - m) + 8}}{5}}
\end{array}\]
Here OD = m. therefore,
\[\begin{array}{*{20}{c}}
{ \Rightarrow m}& = &{\dfrac{{3( - m) - 4( - m) + 8}}{5}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow m}& = &2
\end{array}\]
Now the center of the circle is O (-2, -2) and the radius of the circle is also 2.
Therefore,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - {x_1}} \right)}^2} + {{\left( {y - {y_1}} \right)}^2}}& = &{{R^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x + 2} \right)}^2} + {{\left( {y + 2} \right)}^2}}& = &4
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {y^2} + 4x + 4y + 4}& = &0
\end{array}\]
So, the equation of the circle is ${x^2} + {y^2} + 4x + 4y + 4 = 0$
Note: The first point is to keep in mind that if the circle touches both the axis, then the distance of the center from the X and Y axis will be equal.
Formula Used: \[\begin{array}{*{20}{c}}
d& = &{\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}}
\end{array}\]
Complete step by step solution: The circle lies in the third quadrant. So, the coordinates of the center of the circle will be negative. If the circle touches both the axis, then the distance of the center point from the X and Y axis will be equal.
Let us assume that the coordinate of the circle is O(-m, -m).
According to the given problem, we will draw a figure,

Figure 1
In this problem, we have given an equation 3x – 4y + 8 = 0. Therefore, we will apply the formula of perpendicular distance between the points and the line.
Now,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}}
\end{array}\]
Therefore, we will put the values in the above equation
\[\begin{array}{*{20}{c}}
{ \Rightarrow OD}& = &{\dfrac{{3( - m) - 4( - m) + 8}}{{\sqrt {9 + 16} }}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow OD}& = &{\dfrac{{3( - m) - 4( - m) + 8}}{5}}
\end{array}\]
Here OD = m. therefore,
\[\begin{array}{*{20}{c}}
{ \Rightarrow m}& = &{\dfrac{{3( - m) - 4( - m) + 8}}{5}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow m}& = &2
\end{array}\]
Now the center of the circle is O (-2, -2) and the radius of the circle is also 2.
Therefore,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - {x_1}} \right)}^2} + {{\left( {y - {y_1}} \right)}^2}}& = &{{R^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x + 2} \right)}^2} + {{\left( {y + 2} \right)}^2}}& = &4
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {y^2} + 4x + 4y + 4}& = &0
\end{array}\]
So, the equation of the circle is ${x^2} + {y^2} + 4x + 4y + 4 = 0$
Note: The first point is to keep in mind that if the circle touches both the axis, then the distance of the center from the X and Y axis will be equal.
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