Question

Find the equations of the circles which touch the lines \[3x - 4y + 1 = 0\] and \[4x + 3y - 7 = 0\] and pass through \[\left( {2,3} \right)\].
A. \[{\left( {x - \dfrac{6}{5}} \right)^2} + {\left( {y - \dfrac{{12}}{5}} \right)^2} = 1,{\rm{ }}{\left( {x - 2} \right)^2} + {\left( {y - 8} \right)^2} = 25\]
B. \[{\left( {x + \dfrac{6}{5}} \right)^2} + {\left( {y + \dfrac{{12}}{5}} \right)^2} = 1,{\rm{ }}{\left( {x - 2} \right)^2} + {\left( {y - 8} \right)^2} = 25\]
C. \[{\left( {x + \dfrac{6}{5}} \right)^2} + {\left( {y + \dfrac{{12}}{5}} \right)^2} = 1,{\rm{ }}{\left( {x + 2} \right)^2} + {\left( {y + 8} \right)^2} = 25\]
D. \[{\left( {x - \dfrac{6}{5}} \right)^2} + {\left( {y - \dfrac{{12}}{5}} \right)^2} = 1,{\rm{ }}{\left( {x + 2} \right)^2} + {\left( {y + 8} \right)^2} = 25\]

Answer 1

Hint: In order to solve the question, first using the perpendicular distance formula find the value of r. Next using the distance formula between two points find the value of r. Substitute the two equations and find the required equations of the circle.

Formula Used: The perpendicular distance formula is given as
\[d = \dfrac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}\]
The distance formula between two points is
\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
The general form of the equation of a circle is
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]

Complete step by step solution: Given the lines
\[3x - 4y + 1 = 0\] and \[4x + 3y - 7 = 0\]
Suppose that the center of the circle is (h, k). So here the radius will be the perpendicular distance from the (h, k) to each line.

We know the perpendicular distance formula,
\[d = \dfrac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}\]
That is
\[r = \dfrac{{3h - 4k + 1}}{{\sqrt {{3^2} + {4^2}} }} = \dfrac{{4h + 3k - 7}}{{\sqrt {{4^2} + {3^2}} }}\]
\[r = \dfrac{{3h - 4k + 1}}{{\sqrt {25} }} = \dfrac{{4h + 3k - 7}}{{\sqrt {25} }}\]
\[r = \dfrac{{3h - 4k + 1}}{5} = \dfrac{{4h + 3k - 7}}{5}\]
Squaring on both the sides we get
\[{r^2} = \dfrac{{{{\left( {3h - 4k + 1} \right)}^2}}}{{25}} = \dfrac{{{{\left( {4h + 3k - 7} \right)}^2}}}{{25}}\]. . . . . . (1)
Here also given that (2, 3) is a point on the circle. So using the distance formula between two points
\[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
We can write that
\[r = \sqrt {{{\left( {h - 2} \right)}^2} + {{\left( {k - 3} \right)}^2}} \]
Squaring on both the sides we get
\[{r^2} = {\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2}\]. . . . . . (2)
Substitute equation (2) in equation (1). That is,
\[{\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} = \dfrac{{{{\left( {3h - 4k + 1} \right)}^2}}}{{25}} = \dfrac{{{{\left( {4h + 3k - 7} \right)}^2}}}{{25}}\]
So we can write that
\[{\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} = \dfrac{{{{\left( {3h - 4k + 1} \right)}^2}}}{{25}}\]. . . . . . (3)
And
\[{\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} = \dfrac{{{{\left( {4h + 3k - 7} \right)}^2}}}{{25}}\]. . . . . . (4)
Solving equation (3) and equation (4), we get
\[h = 2,{\rm{ }}k = 8,{\rm{ }}r = 5\]
And
\[h = \dfrac{6}{5},{\rm{ }}k = \dfrac{{12}}{5},{\rm{ }}r = 1\]
We know the general form of equation of circle is
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
Hence the equation of the circle are
\[{\left( {x - 2} \right)^2} + {\left( {y - 8} \right)^2} = 25\]
And
\[{\left( {x - \dfrac{6}{5}} \right)^2} + {\left( {y - \dfrac{{12}}{5}} \right)^2} = 1\]

Option ‘A’ is correct

Note: Students can get confused with perpendicular distance formula and distance formula between two points. The shortest distance from a point to the line is known as the perpendicular distance. Once the coordinates of the center and the radius are found the equation of the circle can be found easily.