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Question: A line \(y=mx+1\) intersects the circle \({{\left( x-3 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=25\)...

A line y=mx+1y=mx+1 intersects the circle (x−3)2+(y+2)2=25{{\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 −35-\dfrac{3}{5} then which of the following options are correct:
(a)6≤m<86\le m < 8
(b) 2≤m<42\le m < 4
(c) 4≤m<64\le m < 6
(d) −3≤m<\-1-3\le m <\- 1

Explanation

Solution

As it is given that line y=mx+1y=mx+1 intersects the circle (x−3)2+(y+2)2=25{{\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+1mx+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+1y=mx+1 which intersects the circle (x−3)2+(y+2)2=25{{\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+1y=mx+1
Now, substituting this value of y in the equation of a circle we get,
(x−3)2+(mx+1+2)2=25 ⇒x2+9−6x+(mx+3)2=25 ⇒x2+9−6x+m2x2+9+6mx=25 \begin{aligned} & {{\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{aligned}
Clubbing the coefficients of x2,x{{x}^{2}},x together we get,
x2(m2+1)+x(6m−6)+18−25=0 ⇒x2(m2+1)+x(6m−6)−7=0 \begin{aligned} & {{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{aligned}
Now, we have given the x coordinates of the midpoint of line segment PQ as −35-\dfrac{3}{5} so the sum of x coordinates of P and Q with divided by 2 is equal to −35-\dfrac{3}{5}.
Sum of x coordinates of PandQ2=−35\dfrac{\text{Sum of x coordinates of PandQ}}{2}=-\dfrac{3}{5}
Multiplying 2 on both the sides of the above equation we get,
Sum of x coordinates of PandQ=−65\text{Sum of x coordinates of PandQ}=-\dfrac{6}{5}
Now, in the above we have derived the quadratic equation in x as:
x2(m2+1)+x(6m−6)−7=0{{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:
−(6m−6)m2+1-\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 −65-\dfrac{6}{5} we get,
−(6m−6)m2+1=−65........Eq.(1) ⇒−6(m−1m2+1)=−65 \begin{aligned} & -\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{aligned}
-6 will be cancelled out from both the sides we get,
m−1m2+1=15\dfrac{m-1}{{{m}^{2}}+1}=\dfrac{1}{5}
On cross multiplication of the above equation we get,
5m−5=m2+1 ⇒m2−5m+6=0 \begin{aligned} & 5m-5={{m}^{2}}+1 \\\ & \Rightarrow {{m}^{2}}-5m+6=0 \\\ \end{aligned}
We are going to solve the above quadratic equation by factorization method as follows:
m2−3m−2m+6=0 ⇒m(m−3)−2(m−3)=0 \begin{aligned} & {{m}^{2}}-3m-2m+6=0 \\\ & \Rightarrow m\left( m-3 \right)-2\left( m-3 \right)=0 \\\ \end{aligned}
Taking m−3m-3 as common we get,
(m−3)(m−2)=0\left( m-3 \right)\left( m-2 \right)=0
Equating (m−3)&(m−2)\left( m-3 \right)\And \left( m-2 \right) to 0 we get,
m−3=0 ⇒m=3 m−2=0 ⇒m=2 \begin{aligned} & m-3=0 \\\ & \Rightarrow m=3 \\\ & m-2=0 \\\ & \Rightarrow m=2 \\\ \end{aligned}
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:
−(6m−6)m2+1=−65-\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(m−1m2+1)=−65-6\left( \dfrac{m-1}{{{m}^{2}}+1} \right)=-\dfrac{6}{5}
-6 will be cancelled out on both the sides we get,
(m−1m2+1)=15\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,

& \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{aligned}$$ 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, $\begin{aligned} & \left( \dfrac{3-1}{{{\left( 3 \right)}^{2}}+1} \right)=\dfrac{1}{5} \\\ & \Rightarrow \dfrac{2}{9+1}=\dfrac{1}{5} \\\ & \Rightarrow \dfrac{2}{10}=\dfrac{1}{5} \\\ & \Rightarrow \dfrac{1}{5}=\dfrac{1}{5} \\\ \end{aligned}$ Hence, L.H.S is equal to R.H.S so the value of m equals 3 satisfying the eq. (1).