Locus is[tex]E_{2}[/tex] = [tex]\sf{x² + y² - x - 2y + 1 = 0}[/tex]
[tex]\\[/tex]
Now after checking options (b)and (d)are correct.
[tex]\rule{200pt}{4pt}[/tex]
Verification :-
Option (b)
Also , [tex]\dfrac{4}{5} \:\:\dfrac{7}{5}[/tex] satisfies [tex]E_{2}[/tex] but again in the case one end of the chord would be (-2,7)which is not included in [tex]E_{1}[/tex] Therefore,[tex]\dfrac{4}{5} \:\:\:\dfrac{7}{5}[/tex] does not lie in [tex]E_{2}[/tex]
Option (d)
(0, [tex]\dfrac{3}{2}[/tex] does not satisfy [tex]E_{1}[/tex] so,it does not lie in [tex]E_{1}[/tex]
Answers & Comments
Verified answer
Step-by-step explanation:
Construction :- Through M, draw a common tangent to two circles intersecting tangent T at N.
Now, NP and NM are tangents to a circle [tex]S_1 [/tex].
We know,
Length of tangents are equal drawn from external point.
[tex]\sf\implies \: NP = NM - - - (1) \\ \\ [/tex]
Also, NM and NQ are tangents to a circle [tex]S_2[/tex].
[tex]\sf\implies \: NM = NQ- - - (2) \\ \\ [/tex]
From equation (1) and (2), we concluded that
[tex]\sf \: NP = NQ \\ \\ [/tex]
[tex]\sf\implies \: N \: is \: the \: midpoint \: of \: PQ \\ \\ [/tex]
Using Midpoint Formula, Coordinates of Q = (0, 1)
Let assume that coordinates of M be (x, y).
Now, we have from equation (1),
[tex]\sf \: NP = NM \\ \\ [/tex]
[tex]\sf \: NP^{2} = NM^{2} \\ \\ [/tex]
[tex]\sf \: {(x - 0)}^{2} + {(y - 1)}^{2} = {( - 2 - 0)}^{2} + {(7 - 1)}^{2} \\ \\ [/tex]
[tex]\sf \: {x}^{2} + {(y - 1)}^{2} = 4 + 36 \\ \\ [/tex]
[tex]\sf \: {x}^{2} + {(y - 1)}^{2} = 40 \\ \\ [/tex]
[tex]\sf\implies {x}^{2} + {(y - 1)}^{2} = {(2 \sqrt{10}) }^{2} \\ \\ [/tex]
So, Locus of point M is circle of centre (0, 1) and radius of circle is [tex]2 \sqrt{10}[/tex]
So,
[tex]\sf\implies E_1 : {x}^{2} + {(y - 1)}^{2} = 40 \\ \\ [/tex]
Now, Consider the point (- 2, 7)
[tex]\sf\implies E_1 : {( - 2)}^{2} + {(7 - 1)}^{2} = 40 \\ \\ [/tex]
[tex]\sf\implies E_1 : 4 + 36 = 40 \\ \\ [/tex]
[tex]\sf\implies E_1 : 40 = 40 \\ \\ [/tex]
[tex]\sf\implies ( - 2,7) \: lies \: on \: E_1 \\ \\ [/tex]
[tex]\bf\implies \: (A) \: option \: is \: wrong. \\ \\ [/tex]
Now, Consider [tex](0, \frac{3}{2}) [/tex]
[tex]\sf\implies E_1 : {(0)}^{2} + {\bigg( \frac{3}{2} - 1\bigg)}^{2} = 40 \\ \\ [/tex]
[tex]\sf\implies E_1 : \frac{1}{4} < 40 \\ \\ [/tex]
[tex]\sf\implies \bigg(0,\dfrac{3}{2}\bigg) \: lies \: in \: E_1 \\ \\ [/tex]
[tex]\bf\implies \: (D) \: option \: is \: wrong. \\ \\ [/tex]
Now, Locus of points in [tex]E_2 [/tex] is a circle with end points of diameter as (0, 1) and (1, 1)
So, equation of circle is given by
[tex]\sf \: (x - 0)(x - 1) + (y - 1)(y -1) = 0 \\ \\ [/tex]
[tex]\sf \: x(x - 1) + {(y - 1)}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: {x}^{2} - x + {(y - 1)}^{2} = 0 \\ \\ [/tex]
[tex]\sf \: {\bigg(x - \dfrac{1}{2}\bigg) }^{2} + {(y - 1)}^{2} = \dfrac{1}{4} \\ \\ [/tex]
So,
[tex]\sf\implies E_2 : {\bigg(x - \dfrac{1}{2} \bigg) }^{2} + {(y - 1)}^{2} = \dfrac{1}{4} \\ \\ [/tex]
Now, Consider [tex](\frac{4}{5}, \frac{7}{5}) [/tex]
[tex]\sf\implies E_2 : {\bigg( \frac{4}{5} - \dfrac{1}{2} \bigg) }^{2} + {\bigg(\dfrac{7}{5} - 1\bigg)}^{2} = \dfrac{1}{4} \\ \\ [/tex]
[tex]\sf\implies E_2 : \frac{9}{100} + \frac{4}{25} = \dfrac{1}{4} \\ \\ [/tex]
[tex]\sf\implies E_2 : \frac{9}{100} + \frac{16}{100} = \dfrac{1}{4} \\ \\ [/tex]
[tex]\sf\implies E_2 : \frac{25}{100} = \dfrac{1}{4} \\ \\ [/tex]
[tex]\sf\implies E_2 : \frac{1}{4} = \dfrac{1}{4} \\ \\ [/tex]
[tex]\sf\implies \bigg(\dfrac{4}{5} ,\dfrac{7}{5} \bigg) \: lies \: on \: E_2 \\ \\ [/tex]
[tex]\sf\implies \bigg(\dfrac{4}{5} ,\dfrac{7}{5} \bigg) \: donot \: lies \: in \: E_2 \\ \\ [/tex]
[tex]\bf\implies \: (B) \: option \: is \: correct. \\ \\ [/tex]
Now, Consider [tex](\frac{1}{2}, 1) [/tex]
[tex]\sf\implies E_2 : {\bigg( \frac{1}{2} - \dfrac{1}{2} \bigg) }^{2} + {\bigg(1 - 1\bigg)}^{2} = \dfrac{1}{4} \\ \\ [/tex]
[tex]\sf\implies E_2 : 0 < \dfrac{1}{4} \\ \\ [/tex]
[tex]\sf\implies \bigg(\dfrac{1}{2} ,1\bigg) \: lies \: in \: E_2 \\ \\ [/tex]
[tex]\bf\implies \: (C) \: option \: is \: correct. \\ \\ [/tex]
[tex]\rule{200pt}{4pt}[/tex]
[tex]\underline{\underline{\bf{Solution : -}}}[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\hookrightarrow{MN = NP = NQ}[/tex]
Locus of M is circle having PQ as it's diameter of circle.
Equation of Circle :
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{(x -2) (x +2)+ (y + 5) (y -7) = 0}[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{x² + y² - 2y - 39 = 0 }[/tex]
[tex]\\[/tex]
Hence ,
[tex]E_{1}[/tex][tex]{x}^{2} + {y}^{2} - 2y \: - 39 = 0[/tex]
Locus of mid point of chord (h,k) of the circle [tex]E_{1}[/tex] is
[tex]\\[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{xh + yk - (y + k) - 39}[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{h² + K² - (y -2k - 39}[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{xh + yk - y - k = h² + k² - 2k}[/tex]
[tex]\\[/tex]
Since, chord is passing through (1,1)
Locus of mid point of chord h,k is
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{h² + K² - (y -2k - 39}[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf\implies{h²+ K - 1 - k = h² + k² - 2k}[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\:\:\:\:\:\:[/tex] [tex]\sf{h² + k² - 2k - h + 1 = 0}[/tex]
[tex]\\[/tex]
Locus is[tex]E_{2}[/tex] = [tex]\sf{x² + y² - x - 2y + 1 = 0}[/tex]
[tex]\\[/tex]
Now after checking options (b)and (d) are correct.
[tex]\rule{200pt}{4pt}[/tex]
Verification : -
Option (b)
Also , [tex]\dfrac{4}{5} \:\:\dfrac{7}{5}[/tex] satisfies [tex]E_{2}[/tex] but again in the case one end of the chord would be (-2,7) which is not included in [tex]E_{1}[/tex] Therefore, [tex]\dfrac{4}{5} \:\:\:\dfrac{7}{5}[/tex] does not lie in [tex]E_{2}[/tex]
Option (d)
(0, [tex]\dfrac{3}{2}[/tex] does not satisfy [tex]E_{1}[/tex] so, it does not lie in [tex]E_{1}[/tex]
[tex]\rule{200pt}{4pt}[/tex]