Question

Find the equation of the circle whose center is (3, -2) and which cut off an intercept of length 6 units on the line \[4x-3y+2=0\].

Answer 1

Hint: The center of the circle is O (3, -2). Use the property that the perpendicular drawn from the center to the chord of a circle bisects the chord and get the length of AP where the equation of chord AB is \[4x-3y+2=0\] . Now, use the formula, Distance = \[\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\] and calculate the perpendicular distance OP. Calculate the length of OA by applying the Pythagoras theorem in \[\Delta OPA\] . Now, solve it further and get the required equation of the circle by using the standard equation of the circle having its center at \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[r\] as the radius,
\[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}\].

Complete step-by-step solution
According to the question, we are given a circle having (3, -2) as its center and cut off an intercept of length 6 units on the line \[4x-3y+2=0\].
The coordinate of the center of the circle, O = (3, -2) ………………………………………….(1)
The length of the intercept = 6 units ……………………………………………..(2)
The equation of the straight line is \[4x-3y+2=0\] ………………………………………..(3)
We know the property that the perpendicular drawn from the center to the chord of a circle bisects the chord.
Using the above property we can say that the perpendicular OP drawn from the center (3, -2) is bisecting the chord AB, \[4x-3y+2=0\].

AP = PB = 3 units ………………………………(4)
We know the formula for perpendicular distance of a point \[\left( {{x}_{1}},{{y}_{1}} \right)\] for the straight line
\[ax+by+c=0\] , Distance = \[\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\] ……………………………………(5)
Now, from equation (1), equation (3), and equation (4), we get
The perpendicular distance, OP = \[\left| \dfrac{\left( 4\times 3 \right)-\left( -2\times 3 \right)+2}{\sqrt{{{4}^{2}}+{{3}^{2}}}} \right|=\left| \dfrac{12+6+2}{\sqrt{25}} \right|=\left| \dfrac{20}{5} \right|=4\] units …………………………………………….(6)
In \[\Delta OPA\] , we have
Perpendicular = OP = 4 units (from equation (6)) ………………………………………(7)
Base = AP = 3 units (from equation (4)) …………………………………..(8)
Hypotenuse = OA ………………………………………….(9)
We know the Pythagoras theorem,
\[{{\left( Hypotenuse \right)}^{2}}={{\left( Base \right)}^{2}}+{{\left( Perpendicular \right)}^{2}}\] …………………………………………….(10)
 Now, from equation (7), equation (8), equation (9), and equation (10), we get
\[\begin{align}
  & \Rightarrow {{\left( OA \right)}^{2}}={{\left( AP \right)}^{2}}+{{\left( OP \right)}^{2}} \\
 & \Rightarrow {{\left( OA \right)}^{2}}={{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}} \\
 & \Rightarrow OA=\sqrt{9+16} \\
\end{align}\]
\[\Rightarrow OA=5\] …………………………………(11)
In the figure above, we can observe that OA is the radius of the required circle.
We also know the standard equation of the circle having its center at \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[r\] as the radius,
\[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}\] ……………………………………………..(12)
From equation (1), equation (11), and equation (12), we get
\[\begin{align}
  & \Rightarrow {{\left( x-3 \right)}^{2}}+{{\left\{ y-\left( -2 \right) \right\}}^{2}}={{5}^{2}} \\
 & \Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=25 \\
\end{align}\]
Hence, the required equation of the circle is \[{{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=25\].

Note: For this type of question, we have to consider two main points. The first one is that perpendicular drawn from the center of the circle to the chord bisects the chord. The second point is the perpendicular distance of a point \[\left( {{x}_{1}},{{y}_{1}} \right)\] for the straight line \[ax+by+c=0\] , Distance = \[\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}+c}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\].