
The equation of circle whose centre is \[(1, - 3)\] and which touches the line \[2{\rm{ }}x - y - 4 = 0\], is
A. \[5{x^2} + 5{y^2} + 10x + 30y + 49 = 0\]
B. \[5{x^2} + 5{y^2} + 10x - 30y - 49 = 0\]
C. \[5{x^2} + 5{y^2} - 10x + 30y - 49 = 0\]
D. None of these
Answer
162.9k+ views
Hint:To answer this question, we will assume that \[{x^2} + {y^2} - 2gx - 2fy + c = 0\] is the equation of the required circle. Following that, we will form equations between \[g,f,c\] and solve them to find their values using the given data and hence find the required equation of a circle with centre \[(1, - 3)\].
Complete step by step Solution:
We have been given that the circle passes through\[(1, - 3)\] and touches the equation
\[2{\rm{ }}x - y - 4 = 0\]
The general equation of a circle is
\[{(x - h)^2} + {(y - k)^2} = {r^2}\].
Here,\[(h,k)\]is the centre of the circle
That is the centre of the circle is: \[(1, - 3)\].
It is given that circle with the centre \[(1, - 3)\] touches the straight line \[2x - y - 4 = 0\].
As a result, the length of the perpendicular dropped from centre \[\left( {1, - 3} \right)\] to the line.
Then \[2x - y - 4 = 0\] will be equal to the radius of the circle.
Radius \[ = \dfrac{{|2 \times 1 + 3 - 4|}}{{\sqrt {{2^2} + {{( - 1)}^2}} }}\]
First we have to simplify the numerator: \[ = \dfrac{1}{{\sqrt {{2^2} + {{( - 1)}^2}} }}\]
Now simplify the denominator\[{2^2} + {( - 1)^2} = 5\]: \[ = \dfrac{1}{{\sqrt 5 }}\]
Hence the equation of the circle is
\[{({\rm{x}} - 1)^2} + {({\rm{y}} - ( - 3))^2} = {\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2}\]
Multiply the minus sign with the number inside the parentheses:
\[{(x - 1)^2} + {(y + 3)^2} = {\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2}\]
Expand the above expression to make to less complicated:
\[ \Rightarrow {{\rm{x}}^2} - 2{\rm{x}} + 1 + {{\rm{y}}^2} + 6{\rm{y}} + 9 = \dfrac{1}{5}\]
Add the constants to make it simpler:
\[ \Rightarrow {{\rm{x}}^2} + {{\rm{y}}^2} - 2{\rm{x}} + 6{\rm{y}} + 10 = \dfrac{1}{5}\]
Multiply\[{\rm{5}}\]with equation on the left side:
\[ \Rightarrow 5{{\rm{x}}^2} + 5{{\rm{y}}^2} - 10{\rm{x}} + 30{\rm{y}} + 50 - 1 = 0\]
Simplify the like terms:
\[\therefore 5{{\rm{x}}^2} + 5{{\rm{y}}^2} - 10{\rm{x}} + 30{\rm{y}} + 49 = 0\]
Therefore,\[5{{\rm{x}}^2} + 5{{\rm{y}}^2} - 10{\rm{x}} + 30{\rm{y}} + 49 = 0\]is the required equation of the circle.
Therefore, the correct option is (D).
Note: The student should start with the basic equation of the circle. After that, try to correctly form the equations based on the data provided in the question. In order to get the correct answer, students should avoid making calculation errors while solving.
Complete step by step Solution:
We have been given that the circle passes through\[(1, - 3)\] and touches the equation
\[2{\rm{ }}x - y - 4 = 0\]
The general equation of a circle is
\[{(x - h)^2} + {(y - k)^2} = {r^2}\].
Here,\[(h,k)\]is the centre of the circle
That is the centre of the circle is: \[(1, - 3)\].
It is given that circle with the centre \[(1, - 3)\] touches the straight line \[2x - y - 4 = 0\].
As a result, the length of the perpendicular dropped from centre \[\left( {1, - 3} \right)\] to the line.
Then \[2x - y - 4 = 0\] will be equal to the radius of the circle.
Radius \[ = \dfrac{{|2 \times 1 + 3 - 4|}}{{\sqrt {{2^2} + {{( - 1)}^2}} }}\]
First we have to simplify the numerator: \[ = \dfrac{1}{{\sqrt {{2^2} + {{( - 1)}^2}} }}\]
Now simplify the denominator\[{2^2} + {( - 1)^2} = 5\]: \[ = \dfrac{1}{{\sqrt 5 }}\]
Hence the equation of the circle is
\[{({\rm{x}} - 1)^2} + {({\rm{y}} - ( - 3))^2} = {\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2}\]
Multiply the minus sign with the number inside the parentheses:
\[{(x - 1)^2} + {(y + 3)^2} = {\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2}\]
Expand the above expression to make to less complicated:
\[ \Rightarrow {{\rm{x}}^2} - 2{\rm{x}} + 1 + {{\rm{y}}^2} + 6{\rm{y}} + 9 = \dfrac{1}{5}\]
Add the constants to make it simpler:
\[ \Rightarrow {{\rm{x}}^2} + {{\rm{y}}^2} - 2{\rm{x}} + 6{\rm{y}} + 10 = \dfrac{1}{5}\]
Multiply\[{\rm{5}}\]with equation on the left side:
\[ \Rightarrow 5{{\rm{x}}^2} + 5{{\rm{y}}^2} - 10{\rm{x}} + 30{\rm{y}} + 50 - 1 = 0\]
Simplify the like terms:
\[\therefore 5{{\rm{x}}^2} + 5{{\rm{y}}^2} - 10{\rm{x}} + 30{\rm{y}} + 49 = 0\]
Therefore,\[5{{\rm{x}}^2} + 5{{\rm{y}}^2} - 10{\rm{x}} + 30{\rm{y}} + 49 = 0\]is the required equation of the circle.
Therefore, the correct option is (D).
Note: The student should start with the basic equation of the circle. After that, try to correctly form the equations based on the data provided in the question. In order to get the correct answer, students should avoid making calculation errors while solving.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
