Answer
384.3k+ views
Hint: Here, we have to find the locus of the midpoint of \[AB\]. We will use the equation of the line and the chord to solve the question. A chord of a circle is a straight line segment whose endpoints both lie on the circle.
Formula used:
We will use the following formulas:
1. The equation of the chord of the circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]bisected at the point \[({x_1},{y_1})\] is given by \[T = {S_1}\]
2. If two lines are perpendicular to each other, we have \[{m_1}{m_2} = \dfrac{a}{b} = - 1\].
Complete Step by Step Solution:
We will first draw a circle with two perpendicular chords.
Here, AB and CD are two perpendicular chords of a circle.
We are given that two perpendicular chords of the circle \[r = a\cos \theta + b\sin \theta \] passing through origin
\[r = acos\theta + bsin\theta \] ……………………………… (1)
Multiplying equation (1) by \[r\] on both sides, we get
\[ \Rightarrow {r^2} = arcos\theta + brsin\theta \]…………………… (2)
The Cartesian equation of a circle is written as \[{r^2} = {x^2} + {y^2}\].
Polar coordinates of a circle are written as \[rcos\theta = x\] and \[rsin\theta = y\].
By substituting the Cartesian equation and rewriting the polar coordinates in equation (2), we have
\[ \Rightarrow {x^2} + {y^2} = ax + by\]
Rewriting the equation , we have
\[ \Rightarrow {x^2} + {y^2} - ax - by = 0\]
Thus the equation of the circle \[{x^2} + {y^2} - ax - by = 0\]
Now, we have to find the locus of the midpoint of the chord\[AB\].
Let the midpoint of chord \[AB\]be\[\left( {h,k} \right)\]
Since we have to find the locus at Midpoint of the chord AB at \[\left( {h,k} \right)\].
\[hx + ky - \dfrac{{a\left( {x + h} \right)}}{2} - \dfrac{{b\left( {y + k} \right)}}{2} = {h^2} + {k^2} - ah - bk\]
\[ \Rightarrow hx + ky - \dfrac{{ax}}{2} - \dfrac{{by}}{2} - {h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2} = 0\].
Homogenizing the equation, we get
\[ \Rightarrow {x^2} + {y^2} - \left( {ax + by} \right)\dfrac{{(hx + ky - \dfrac{{ax}}{2} - \dfrac{{by}}{2})}}{{({h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2})}} = 0\]
Rewriting the equation, we get
\[ \Rightarrow {x^2}\left( {{h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2}} \right) + {y^2}\left( {{h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2}} \right) - \left( {ah{x^2} - \dfrac{{{a^2}{x^2}}}{2} + bk{y^2} - \dfrac{{{b^2}{y^2}}}{2} + xy\left( p \right)} \right) = 0\]
Here \[p\] is the coefficient of \[xy\].
Since the two chords are perpendicular to each other, we have
\[ \Rightarrow {m_1}{m_2} = \dfrac{a}{b} = - 1\]
\[ \Rightarrow {h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2} = - {h^2} - {k^2} + \dfrac{{ah}}{2} + \dfrac{{bk}}{2}\]
\[ \Rightarrow 2\left( {{h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2}} \right) = 0\]
Rewriting the equation, we get
\[ \Rightarrow {h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2} = 0\]
Re-substituting \[(h,k)\] as \[(x,y)\], we get
Locus of the Midpoint is \[{x^2} + {y^2} - \dfrac{{ax}}{2} - \dfrac{{by}}{2} = 0\]
Therefore, the locus of the midpoint is \[{x^2} + {y^2} = \dfrac{{ax}}{2} + \dfrac{{by}}{2}\].
Hence option A is the correct answer.
Note:
Homogenization is the process of making it homogenous. That is, we should make the degree of every term the same. We have to be clear that the locus should be found out for the midpoint of the chord. When a point moves in a plane according to some given conditions the path along which it moves is called a locus. We should notice whether the points lie inside or outside or on the circle.
Formula used:
We will use the following formulas:
1. The equation of the chord of the circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]bisected at the point \[({x_1},{y_1})\] is given by \[T = {S_1}\]
2. If two lines are perpendicular to each other, we have \[{m_1}{m_2} = \dfrac{a}{b} = - 1\].
Complete Step by Step Solution:
We will first draw a circle with two perpendicular chords.
![seo images](https://www.vedantu.com/question-sets/7bb41428-b56f-496a-852c-85de8db0e6671055995471051777767.png)
Here, AB and CD are two perpendicular chords of a circle.
We are given that two perpendicular chords of the circle \[r = a\cos \theta + b\sin \theta \] passing through origin
\[r = acos\theta + bsin\theta \] ……………………………… (1)
Multiplying equation (1) by \[r\] on both sides, we get
\[ \Rightarrow {r^2} = arcos\theta + brsin\theta \]…………………… (2)
The Cartesian equation of a circle is written as \[{r^2} = {x^2} + {y^2}\].
Polar coordinates of a circle are written as \[rcos\theta = x\] and \[rsin\theta = y\].
By substituting the Cartesian equation and rewriting the polar coordinates in equation (2), we have
\[ \Rightarrow {x^2} + {y^2} = ax + by\]
Rewriting the equation , we have
\[ \Rightarrow {x^2} + {y^2} - ax - by = 0\]
Thus the equation of the circle \[{x^2} + {y^2} - ax - by = 0\]
Now, we have to find the locus of the midpoint of the chord\[AB\].
Let the midpoint of chord \[AB\]be\[\left( {h,k} \right)\]
Since we have to find the locus at Midpoint of the chord AB at \[\left( {h,k} \right)\].
\[hx + ky - \dfrac{{a\left( {x + h} \right)}}{2} - \dfrac{{b\left( {y + k} \right)}}{2} = {h^2} + {k^2} - ah - bk\]
\[ \Rightarrow hx + ky - \dfrac{{ax}}{2} - \dfrac{{by}}{2} - {h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2} = 0\].
Homogenizing the equation, we get
\[ \Rightarrow {x^2} + {y^2} - \left( {ax + by} \right)\dfrac{{(hx + ky - \dfrac{{ax}}{2} - \dfrac{{by}}{2})}}{{({h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2})}} = 0\]
Rewriting the equation, we get
\[ \Rightarrow {x^2}\left( {{h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2}} \right) + {y^2}\left( {{h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2}} \right) - \left( {ah{x^2} - \dfrac{{{a^2}{x^2}}}{2} + bk{y^2} - \dfrac{{{b^2}{y^2}}}{2} + xy\left( p \right)} \right) = 0\]
Here \[p\] is the coefficient of \[xy\].
Since the two chords are perpendicular to each other, we have
\[ \Rightarrow {m_1}{m_2} = \dfrac{a}{b} = - 1\]
\[ \Rightarrow {h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2} = - {h^2} - {k^2} + \dfrac{{ah}}{2} + \dfrac{{bk}}{2}\]
\[ \Rightarrow 2\left( {{h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2}} \right) = 0\]
Rewriting the equation, we get
\[ \Rightarrow {h^2} + {k^2} - \dfrac{{ah}}{2} - \dfrac{{bk}}{2} = 0\]
Re-substituting \[(h,k)\] as \[(x,y)\], we get
Locus of the Midpoint is \[{x^2} + {y^2} - \dfrac{{ax}}{2} - \dfrac{{by}}{2} = 0\]
Therefore, the locus of the midpoint is \[{x^2} + {y^2} = \dfrac{{ax}}{2} + \dfrac{{by}}{2}\].
Hence option A is the correct answer.
Note:
Homogenization is the process of making it homogenous. That is, we should make the degree of every term the same. We have to be clear that the locus should be found out for the midpoint of the chord. When a point moves in a plane according to some given conditions the path along which it moves is called a locus. We should notice whether the points lie inside or outside or on the circle.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)