Solveeit Logo
Login

Question

Question: Let T be the line passing through the points P(-2, 7) and Q(2, -5). Let \({{F}_{1}}\) be the set of ...

Let T be the line passing through the points P(-2, 7) and Q(2, -5). Let F1{{F}_{1}} be the set of all pair of circles (S1,S2)\left( {{S}_{1}},{{S}_{2}} \right) such that T is the tangent to S1{{S}_{1}} at P and tangent to S2{{S}_{2}}at Q, and also such that S1{{S}_{1}} and S2{{S}_{2}} touch each other at point, say M. Let E1{{E}_{1}} be the set representing the locus of M as the pair (S1,S2)\left( {{S}_{1}},{{S}_{2}} \right) varies in F1{{F}_{1}}. Let the set of all straight line segments joining a pair of distinct points of E1{{E}_{1}} and passing through the point R(1, 1) be F2{{F}_{2}}. Let E2{{E}_{2}} be the set of midpoints of the line segments in the set F2{{F}_{2}}. Then, which one of the following statements is/are true?
(a) The point (-2, 7) lies inE1{{E}_{1}}
(b) The point (45,75)\left( \dfrac{4}{5},\dfrac{7}{5} \right)does not lie in E2{{E}_{2}}
(c) The point (12,1)\left( \dfrac{1}{2},1 \right) lies in E2{{E}_{2}}
(d) The point (0,32)\left( 0,\dfrac{3}{2} \right) does not lie inE1{{E}_{1}}

Explanation

Solution

First, before proceeding with this, we must draw a general diagram of the given condition so that we can understand the question easily by considering C1,r1{{C}_{1}},{{r}_{1}} being the center and radius of the first circle and C2,r2{{C}_{2}},{{r}_{2}} being the center and radius of the second circle. Then, by using the angle sum property of the triangle which gives that sum of all the angles of the triangle is 180∘{{180}^{\circ }} for triangle PMQ. Then, we get the angle ∠PMQ\angle PMQ as 90∘{{90}^{\circ }} which means that triangle PMQ forms a semicircle with center at M and diameter points as P and Q and by using the equation of circle by diameter points as we get the desired result.

Complete step-by-step solution:
In this question, we are supposed to find the correct statements when T is the line passing through the points P(-2, 7) and Q(2, -5). Let F1{{F}_{1}} be the set of all pair of circles (S1,S2)\left( {{S}_{1}},{{S}_{2}} \right) such that T is the tangent to S1{{S}_{1}} at P and tangent to S2{{S}_{2}} at Q, and also such that S1{{S}_{1}} and S2{{S}_{2}} touch each other at point, say M. Let E1{{E}_{1}} be the set representing the locus of M as the pair (S1,S2)\left( {{S}_{1}},{{S}_{2}} \right)varies in F1{{F}_{1}}. Let the set of all straight line segments joining a pair of distinct points of E1{{E}_{1}} and passing through the point R(1, 1) be F2{{F}_{2}}. Let E2{{E}_{2}} be the set of midpoints of the line segments in the set F2{{F}_{2}}.
So, before proceeding with this, we must draw a general diagram of the given condition so that we can understand the question easily by considering C1,r1{{C}_{1}},{{r}_{1}} being the center and radius of the first circle and C2,r2{{C}_{2}},{{r}_{2}} being the center and radius of the second circle as:

Also, we have considered the angle made by PM line in the triangle PMC1PM{{C}_{1}} as θ\theta and similarly angle made by QM in the triangle QMC2QM{{C}_{2}} be ϕ\phi .
Then, we can see that in triangle PMC1PM{{C}_{1}}, sides C1M{{C}_{1}}M and PC1P{{C}_{1}} are the radius of the circle which gives us:
∠PMC1=θ\angle PM{{C}_{1}}=\theta
Similarly, we can see that in triangle QMC2QM{{C}_{2}}, sides C2M{{C}_{2}}M and QC2Q{{C}_{2}} are the radius of the circle which gives us:
∠QMC2=ϕ\angle QM{{C}_{2}}=\phi
Then, we can see clearly that C1C2{{C}_{1}}{{C}_{2}} is the straight line which makes the total angle of 180∘{{180}^{\circ }}.
So, by using this, we get the value of ∠PMQ\angle PMQ as:
∠PMQ+∠PMC1+∠QMC2=180∘ ⇒∠PMQ+θ+ϕ=180∘ ⇒∠PMQ=180∘−θ−ϕ \begin{aligned} & \angle PMQ+\angle PM{{C}_{1}}+\angle QM{{C}_{2}}={{180}^{\circ }} \\\ & \Rightarrow \angle PMQ+\theta +\phi ={{180}^{\circ }} \\\ & \Rightarrow \angle PMQ={{180}^{\circ }}-\theta -\phi \\\ \end{aligned}
Also, we need to get the other two angles of the triangle PMQ by using the figure as:
θ+∠MPQ=90∘ ⇒∠MPQ=90∘−θ \begin{aligned} & \theta +\angle MPQ={{90}^{\circ }} \\\ & \Rightarrow \angle MPQ={{90}^{\circ }}-\theta \\\ \end{aligned}
Similarly, for other angle also by using the figure, we get:
ϕ+∠MQP=90∘ ⇒∠MQP=90∘−ϕ \begin{aligned} & \phi +\angle MQP={{90}^{\circ }} \\\ & \Rightarrow \angle MQP={{90}^{\circ }}-\phi \\\ \end{aligned}
Now, by using the angle sum property of the triangle which gives that sum of all the angles of the triangle is 180∘{{180}^{\circ }}.
So, by applying the same property to triangle PMQ, we get:
∠PMQ+∠PQM+∠MPQ=180∘ ⇒180∘−θ−ϕ+90∘−ϕ+90∘−θ=180∘ ⇒2(θ+ϕ)=180∘ ⇒(θ+ϕ)=90∘ \begin{aligned} & \angle PMQ+\angle PQM+\angle MPQ={{180}^{\circ }} \\\ & \Rightarrow {{180}^{\circ }}-\theta -\phi +{{90}^{\circ }}-\phi +{{90}^{\circ }}-\theta ={{180}^{\circ }} \\\ & \Rightarrow 2\left( \theta +\phi \right)={{180}^{\circ }} \\\ & \Rightarrow \left( \theta +\phi \right)={{90}^{\circ }} \\\ \end{aligned}
So, we get the angle ∠PMQ\angle PMQ as 90∘{{90}^{\circ }} which means that triangle PMQ forms a semicircle with centre at M and diameter points as P and Q.
Now, we also know that the equation of the circle with diameter points (x1,x2)\left( {{x}_{1}},{{x}_{2}} \right) and (y1,y2)\left( {{y}_{1}},{{y}_{2}} \right) is given by :
(x−x1)(x−x2)+(y−y1)(y−y2)=0\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right)=0
So, by using the diameter points P(-2, 7) and Q(2, -5), we get the equation of circle as:
(x+2)(x−2)+(y−7)(y+5)=0 ⇒x2−4+y2−2y−35=0 ⇒x2+y2−2y−39=0 \begin{aligned} & \left( x+2 \right)\left( x-2 \right)+\left( y-7 \right)\left( y+5 \right)=0 \\\ & \Rightarrow {{x}^{2}}-4+{{y}^{2}}-2y-35=0 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-2y-39=0 \\\ \end{aligned}
So, we get the locus of E1{{E}_{1}} by the expression as x2+y2−2y−39=0{{x}^{2}}+{{y}^{2}}-2y-39=0.
Now, E2{{E}_{2}} be the mid points of the points given in the question (1, 1) and the above equation centre (0, 1) and we get the locus of E2{{E}_{2}} as:
(x−1)(x−0)+(y−1)(y−1)=0 ⇒x2−x+y2−2y+1=0 ⇒x2+y2−x−2y+1=0 \begin{aligned} & \left( x-1 \right)\left( x-0 \right)+\left( y-1 \right)\left( y-1 \right)=0 \\\ & \Rightarrow {{x}^{2}}-x+{{y}^{2}}-2y+1=0 \\\ & \Rightarrow {{x}^{2}}+{{y}^{2}}-x-2y+1=0 \\\ \end{aligned}

Hence, options (a) and (d) are correct.

Note: Now, to solve these types of questions we need to know some of the basics of the identities beforehand to solve the expression easily. So, the required expressions are as follows:
(a−b)2=a2+b2−2ab (a−b)(a+b)=a2−b2 \begin{aligned} & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ & \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \\\ \end{aligned}
Also we must know the angle sum property of the triangle which states that sum of all the angle of a triangle is 180∘{{180}^{\circ }} and straight line makes a total angle of ${{180}^{\circ }}.