Question

Find the equation of the circle which touches the line \[4x – 3y + 10 = 0\] and \[4x – 3y – 30 = 0\] and whose centre lies on the line \[2x + y = 0\]

Answer 1

Hint: In this question, we need to find the equation of the circle . Also given that the centre of the circle lies on the line \[2x + y = 0\] . First by using the two given parallel lines, we can find the diameter by using distance between two parallel lines formula. Then by using the diameter, we can find the radius of the circle. As we know that that radius is half of its diameter. Then by using the formula of the perpendicular distance between centre and the line, we can find the centre of the circle. Using this we can find the equation of the circle.

Complete step-by-step answer:

Given , two lines \[4x – 3y + 10 = 0\] and \[4x – 3y – 30 = 0\].
On observing the two given lines, the lines are parallel .
First let us find the diameter of the circle by using the given two parallel lines.
That is \[d = \dfrac{\left| C_{1} – C_{2} \right|}{\sqrt{A^{2} + B^{2}}}\]
Here \[C_{1}\] is \[10\] , \[C_{2}\] is \[- 30\] , \[A\] is \[4\] and \[B\] is \[- 3\] .
Now substituting the known values in the formula,
We get,
\[\Rightarrow \ d = \dfrac{\left| 10 - \left( - 30 \right) \right|}{\sqrt{\left( 4 \right)^{2} + \left( - 3 \right)^{2}}}\]
On simplifying,
We get,
\[d = \dfrac{\left| 10 + 30 \right|}{\sqrt{16 + 9}}\]
On further simplifying,
We get,
\[d = \dfrac{40}{5}\]
On dividing,
We get,
\[d = 8\]
We know that radius is half of its diameter.
That is \[r = \dfrac{d}{2}\]
Thus we get radius as \[4\]
Let the centre of the circle be at \[(h,\ k)\]
Given that the centre lies on the line \[2x + y = 0\].
Now on substituting \[(h,\ k)\] in the place of \[x\] and \[y\],
We get,
\[\Rightarrow \ 2h + k = 0\]
On subtracting both sides by \[2h\] ,
We get,
\[\Rightarrow \ k = - 2h\]
Thus the centre is \[(h,\ - 2h)\] ••• (1)
Now the perpendicular distance between the centre \[(h,\ - 2h)\] from the line \[4x – 3y + 10 = 0\] is \[4\] .
We know that the perpendicular distance from the centre to the line is \[d = \dfrac{\left| Ax_{1} + By_{1} + C \right|}{\sqrt{A^{2} + B^{2}}}\]
On substituting the known values,
We get,
\[4 = \dfrac{\left| 4\left( h \right) + \left( - 3 \right)\left( - 2h \right) + 10 \right|}{\sqrt{4^{2} + \left( - 3 \right)^{2}}}\]
On simplifying,
We get,
\[\dfrac{\left| 4h + 6h + 10 \right|}{\sqrt{16 + 9}} = 4\]
On further simplifying,
We get,
\[\dfrac{\left| 10h + 10 \right|}{5} = 4\]
On taking the common terms outside,
We get,
\[10\dfrac{(h + 1)}{5} = 4\ \]
On simplifying,
We get,
\[2(h + 1) = 4\]
On further simplifying,
We get,
\[2h + 2 = 4\]
On subtracting both sides by \[2\] ,
We get,
\[2h = 2\]
On dividing both sides by \[2\] ,
We get,
\[\Rightarrow \ h = 1\]
On substituting the value of \[h\] in equation (1) ,
We get,
The centre as \[(1,\ - 2)\]
We know that the general equation of the circle is \[\left( x – h \right)^{2} + \left( y – k \right)^{2} = r^{2}\]
Now on substituting the known values,
We get,
\[\left( x – 1 \right)^{2} + \left( y - \left( - 2 \right) \right)^{2} = 4^{2}\]
On simplifying,
We get,
\[\left( x – 1 \right)^{2} + \left( y + 2 \right)^{2} = 16\]
Thus we get the equation of the circle as \[\left( x – 1 \right)^{2} + \left( y + 2 \right)^{2} = 16\].
Final answer :
The equation of the circle is \[\left( x – 1 \right)^{2} + \left( y + 2 \right)^{2} = 16\].

Note: In order to solve these types of questions, we should have a strong grip over equations of the circle. We should be careful while taking the numbers outside the parentheses. Also we should not get confused with the formula of the distance between two parallel lines and perpendicular distance from the centre to the line. We also need to know that mathematically, in a circle when a line a drawn from the centre to any point on the circle then that line is known as the radius of the circle similarly when a line is drawn from a end point to the other end point of the circle passing through the centre of the circle is known as diameter of a circle.