
Circles are drawn through the points \[\left( {2,0} \right)\]to cut the intercept of length \[5\] units on the x-axis. If their center lies in the first quadrant, then find their equation.
A. \[{x^2} + {y^2} - 9x + 2ky + 14 = 0\]
B. \[3{x^2} + 3{y^2} + 27x - 2ky + 42 = 0\]
C. \[{x^2} + {y^2} - 9x - 2ky + 14 = 0\]
D. \[{x^2} + {y^2} - 2kx - 9y + 14 = 0\]
Answer
163.8k+ views
Hint:
In order to solve the question, first find the center of the circle. Then, find the radius of the circle. Finally, using the general form of the equation of the circle, find the required equation.
Formula Used:
Equation of the circle is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
Complete step-by-step answer:
Consider the figure shown below.

Image: The diagram represents the circle
Here the length of AB is 5 units. So the center is
\[ = \left( {\dfrac{{2 + 7}}{2},k} \right)\]
\[ = \left( {\dfrac{9}{2},k} \right)\]
Also, here the radius is
\[r = \sqrt {{{\left( {\dfrac{5}{2}} \right)}^2} + {k^2}} \]
\[r = \sqrt {\dfrac{{25}}{4} + {k^2}} \]
We know that the equation of a circle is given as
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
Where (h, k) is the center and r is the radius.
That is, we can write the equation as
\[{\left( {x - \dfrac{9}{2}} \right)^2} + {\left( {y - k} \right)^2} = \dfrac{{25}}{4} + {k^2}\]
\[{x^2} + \dfrac{{81}}{4} - 9x + {y^2} + {k^2} - 2ky = \dfrac{{25}}{4} + {k^2}\] . . . . . . ( \[ \because {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\])
\[{x^2} - 9x + {y^2} - 2ky + \dfrac{{56}}{4} = 0\]
\[{x^2} + {y^2} - 9x - 2ky + 14 = 0\]
Hence option C is the correct answer.
Note:
Students can get confused while taking the center points. Always remember that the center of a circle is the fixed point in the middle of the circle. Once the coordinates of the center and the radius are found the equation of the circle can be found easily.
In order to solve the question, first find the center of the circle. Then, find the radius of the circle. Finally, using the general form of the equation of the circle, find the required equation.
Formula Used:
Equation of the circle is \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
\[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
Complete step-by-step answer:
Consider the figure shown below.

Image: The diagram represents the circle
Here the length of AB is 5 units. So the center is
\[ = \left( {\dfrac{{2 + 7}}{2},k} \right)\]
\[ = \left( {\dfrac{9}{2},k} \right)\]
Also, here the radius is
\[r = \sqrt {{{\left( {\dfrac{5}{2}} \right)}^2} + {k^2}} \]
\[r = \sqrt {\dfrac{{25}}{4} + {k^2}} \]
We know that the equation of a circle is given as
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\]
Where (h, k) is the center and r is the radius.
That is, we can write the equation as
\[{\left( {x - \dfrac{9}{2}} \right)^2} + {\left( {y - k} \right)^2} = \dfrac{{25}}{4} + {k^2}\]
\[{x^2} + \dfrac{{81}}{4} - 9x + {y^2} + {k^2} - 2ky = \dfrac{{25}}{4} + {k^2}\] . . . . . . ( \[ \because {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\])
\[{x^2} - 9x + {y^2} - 2ky + \dfrac{{56}}{4} = 0\]
\[{x^2} + {y^2} - 9x - 2ky + 14 = 0\]
Hence option C is the correct answer.
Note:
Students can get confused while taking the center points. Always remember that the center of a circle is the fixed point in the middle of the circle. Once the coordinates of the center and the radius are found the equation of the circle can be found easily.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

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

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
