
The equation of the circle passing through $\left( {4,5} \right)$ and having the center $\left( {2,2} \right)$ is
A. ${x^2} + {y^2} + 4x + 4y - 5 = 0$
B. ${x^2} + {y^2} - 4x - 4y - 5 = 0$
C. ${x^2} + {y^2} - 4x = 13$
D. ${x^2} + {y^2} - 4x - 4y - 5 = 0$
Answer
164.4k+ views
Hint: In this question, we are given the coordinates of the center and the point through which the circle is passing. Now, use the general equation of the circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$whose center is $\left( { - g, - f} \right)$. Here, calculate the value of $g$and $f$ using $k = - g$, $h = - f$ where $\left( {h,k} \right)$are $\left( {2,2} \right)$. Now put the value of $x$ , $y$ in the required equation of the circle to get the value of $c$. Lastly, put the value of $c$ in the equation and you’ll get the equation of the circle.
Formula Used:
General equation of circle, ${x^2} + {y^2} + 2gx + 2fy + c = 0$
Complete step by step solution:
Given that,
Center of the circle is $\left( {2,2} \right)$ i.e., $\left( {h,k} \right)$

Image: Circle
The general form of the equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$whose center is $\left( { - g, - f} \right)$
Therefore, $k = - g$, $h = - f$
It implies that, $g = - 2$, $f = - 2$
Now, the equation of the circle is ${x^2} + {y^2} - 4x - 4y + c = 0 - - - - - - \left( 1 \right)$
Also, the equation is passing through the point $\left( {4,5} \right)$, which means points will satisfy the above equation.
Put $x = 4$ and $y = 5$ in equation (1)
${\left( 4 \right)^2} + {\left( 5 \right)^2} - 4\left( 4 \right) - 4\left( 5 \right) + c = 0$
$16 + 25 - 16 - 20 + c = 0$
$c = - 5$
Equation (1) will be,
Equation of the circle is ${x^2} + {y^2} - 4x - 4y - 5 = 0$
Option ‘B’ is correct
Note: To solve such a question, first try to make the figure and understand the question properly. One should always remember each and every equation of the circle (General and standard both). Also, the formulas to find the radius and the distance between any point and line.
Formula Used:
General equation of circle, ${x^2} + {y^2} + 2gx + 2fy + c = 0$
Complete step by step solution:
Given that,
Center of the circle is $\left( {2,2} \right)$ i.e., $\left( {h,k} \right)$

Image: Circle
The general form of the equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$whose center is $\left( { - g, - f} \right)$
Therefore, $k = - g$, $h = - f$
It implies that, $g = - 2$, $f = - 2$
Now, the equation of the circle is ${x^2} + {y^2} - 4x - 4y + c = 0 - - - - - - \left( 1 \right)$
Also, the equation is passing through the point $\left( {4,5} \right)$, which means points will satisfy the above equation.
Put $x = 4$ and $y = 5$ in equation (1)
${\left( 4 \right)^2} + {\left( 5 \right)^2} - 4\left( 4 \right) - 4\left( 5 \right) + c = 0$
$16 + 25 - 16 - 20 + c = 0$
$c = - 5$
Equation (1) will be,
Equation of the circle is ${x^2} + {y^2} - 4x - 4y - 5 = 0$
Option ‘B’ is correct
Note: To solve such a question, first try to make the figure and understand the question properly. One should always remember each and every equation of the circle (General and standard both). Also, the formulas to find the radius and the distance between any point and line.
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

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
