Question

A line $y=mx+1$ intersects the circle ${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=25$ at points P and Q. If the midpoint of the line segment PQ has x coordinate as $-\dfrac{3}{5}$ then which of the following options are correct:
(a)$6\le m < 8$
(b) $2\le m < 4$
(c) $4\le m < 6$
(d) $-3\le m <- 1$

Answer 1

Hint: As it is given that line $y=mx+1$ intersects the circle ${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=25$ at points P and Q so find the intersection of the line with the circle to get the coordinates of points P and Q by substituting the value of y as $mx+1$ in the equation of the circle and solve for x then you will get the quadratic in x. Then the x coordinate of the midpoint of line segment PQ is the sum of the x coordinates of point P and Q and divided by 2 so using the quadratic equation in x we can find the coordinates of the midpoint of PQ and hence, get the relation of m.

Complete step-by-step answer:
We have given a line $y=mx+1$ which intersects the circle ${{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=25$ at points P and Q so we are going to find the coordinates of P and Q by finding the intersection of the line and the circle.
The y in equation of the given line is equal to:
$y=mx+1$
Now, substituting this value of y in the equation of a circle we get,
$\begin{align}
  & {{\left( x-3 \right)}^{2}}+{{\left( mx+1+2 \right)}^{2}}=25 \\
 & \Rightarrow {{x}^{2}}+9-6x+{{\left( mx+3 \right)}^{2}}=25 \\
 & \Rightarrow {{x}^{2}}+9-6x+{{m}^{2}}{{x}^{2}}+9+6mx=25 \\
\end{align}$
Clubbing the coefficients of ${{x}^{2}},x$ together we get,
$\begin{align}
  & {{x}^{2}}\left( {{m}^{2}}+1 \right)+x\left( 6m-6 \right)+18-25=0 \\
 & \Rightarrow {{x}^{2}}\left( {{m}^{2}}+1 \right)+x\left( 6m-6 \right)-7=0 \\
\end{align}$
Now, we have given the x coordinates of the midpoint of line segment PQ as $-\dfrac{3}{5}$ so the sum of x coordinates of P and Q with divided by 2 is equal to $-\dfrac{3}{5}$.
$\dfrac{\text{Sum of x coordinates of PandQ}}{2}=-\dfrac{3}{5}$
 Multiplying 2 on both the sides of the above equation we get,
$\text{Sum of x coordinates of PandQ}=-\dfrac{6}{5}$
Now, in the above we have derived the quadratic equation in x as:
${{x}^{2}}\left( {{m}^{2}}+1 \right)+x\left( 6m-6 \right)-7=0$
The roots of the equation are x coordinates of P and Q and we know that sum of the roots is equal to:
$-\dfrac{\left( 6m-6 \right)}{{{m}^{2}}+1}$
And sum of the roots is equal to the sum of x coordinates of P and Q which we have derived above so equating the above expression to $-\dfrac{6}{5}$ we get,
$\begin{align}
  & -\dfrac{\left( 6m-6 \right)}{{{m}^{2}}+1}=-\dfrac{6}{5}........Eq.(1) \\
 & \Rightarrow -6\left( \dfrac{m-1}{{{m}^{2}}+1} \right)=-\dfrac{6}{5} \\
\end{align}$
-6 will be cancelled out from both the sides we get,
$\dfrac{m-1}{{{m}^{2}}+1}=\dfrac{1}{5}$
On cross multiplication of the above equation we get,
$\begin{align}
  & 5m-5={{m}^{2}}+1 \\
 & \Rightarrow {{m}^{2}}-5m+6=0 \\
\end{align}$
We are going to solve the above quadratic equation by factorization method as follows:
$\begin{align}
  & {{m}^{2}}-3m-2m+6=0 \\
 & \Rightarrow m\left( m-3 \right)-2\left( m-3 \right)=0 \\
\end{align}$
Taking $m-3$ as common we get,
$\left( m-3 \right)\left( m-2 \right)=0$
Equating $\left( m-3 \right)\And \left( m-2 \right)$ to 0 we get,
$\begin{align}
  & m-3=0 \\
 & \Rightarrow m=3 \\
 & m-2=0 \\
 & \Rightarrow m=2 \\
\end{align}$
Hence, we got the values of m as 2 and 3.
These values of m are matching with the option (b) because the range contains both 2 and 3.
Hence, the correct option is (b).

Note:You can check the values of m that you got above by substituting the values of m in eq. (1) and whether they are satisfying that equation or not.
Eq. (1) that we have shown in the above solution is:
$-\dfrac{\left( 6m-6 \right)}{{{m}^{2}}+1}=-\dfrac{6}{5}$
Taking -6 as common in the numerator of the L.H.S of the above equation we get,
$-6\left( \dfrac{m-1}{{{m}^{2}}+1} \right)=-\dfrac{6}{5}$
-6 will be cancelled out on both the sides we get,
$\left( \dfrac{m-1}{{{m}^{2}}+1} \right)=\dfrac{1}{5}$
We got the value of m as 2 and 3 so substituting these values of m one by one in the above equation we get,
Substituting the value of m as 2 we get,
\[\begin{align}
  & \left( \dfrac{2-1}{{{\left( 2 \right)}^{2}}+1} \right)=\dfrac{1}{5} \\
 & \Rightarrow \dfrac{1}{4+1}=\dfrac{1}{5} \\
 & \Rightarrow \dfrac{1}{5}=\dfrac{1}{5} \\
\end{align}\]
Hence, L.H.S is equal to R.H.S so the value of m equals 2 satisfying the eq. (1).
Substituting the value of m as 3 we get,
$\begin{align}
  & \left( \dfrac{3-1}{{{\left( 3 \right)}^{2}}+1} \right)=\dfrac{1}{5} \\
 & \Rightarrow \dfrac{2}{9+1}=\dfrac{1}{5} \\
 & \Rightarrow \dfrac{2}{10}=\dfrac{1}{5} \\
 & \Rightarrow \dfrac{1}{5}=\dfrac{1}{5} \\
\end{align}$
Hence, L.H.S is equal to R.H.S so the value of m equals 3 satisfying the eq. (1).