
A circle $x^{2}+y^{2}+2 g x+2 f y+c=0$ passing through$(4,-2)$ is concentric to the circle $x^{2}+y^{2}-2 x+4 y+20=0$, then the value of $c$ will be
A. -4
B. 0
C. 4
D. 1
Answer
218.7k+ views
Hint: The center of a circle is a location inside the circle that is situated in the middle of the circumference.The radius of a circle is the constant distance from the circle's center to any point on the circle.A circle's diameter is defined as the segment of a line that connects two locations on the circle and passes through its center.
Complete step by step solution:
Since the circle $S_{1} \equiv 3 x^{2}+y^{2}+2 g x+2 f y+c=0$ is concentric to $S_{2} \equiv x^{2}+y^{2}-$ $2 x+4 y+20=0$
$\therefore$ both have the same center.
center of $\mathrm{S}_{2}=\left(\dfrac{-1}{2} \times-2, \dfrac{1}{2} \times 4\right)=(-1,2)$
$\therefore$ for $\mathrm{S}_{1} g=-1 \mathrm{f}=2$
The equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. The coordinates of the circle's center and radius are found using this general form, where g, f, and c are constants. The general form of the equation of a circle makes it difficult to identify any significant properties about any specific circle, in contrast to the standard form, which is simpler to comprehend. So, to quickly change from the generic form to the standard form, we will use the completing the square formula.
$\therefore$ Equation of circle $\mathrm{S}_{1}$ is $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+4 \mathrm{y}+\mathrm{c}=0$ as it passes through $(4,-2)$
$\therefore x^{2}+y^{2}-2 x+4 y+c=0$
$\Rightarrow 4^{2}+(-2)^{2}-2 \times 4+4 \times-2+c=0$
$\Rightarrow 16+4-8-8+c=0$
$\Rightarrow \mathrm{c}=-4$
Option ‘C’ is correct
Note: If two objects in geometry have a common center, they are said to be concentric. Due to their shared center, regular polygons, regular polyhedra, and circles are all concentric. Two circles that are concentric in Euclidean geometry always have different radii but the same center.
Circles having the same or a shared center are known as concentrators. In other terms, circles are said to be concentric if two or more of them share the same center.
Complete step by step solution:
Since the circle $S_{1} \equiv 3 x^{2}+y^{2}+2 g x+2 f y+c=0$ is concentric to $S_{2} \equiv x^{2}+y^{2}-$ $2 x+4 y+20=0$
$\therefore$ both have the same center.
center of $\mathrm{S}_{2}=\left(\dfrac{-1}{2} \times-2, \dfrac{1}{2} \times 4\right)=(-1,2)$
$\therefore$ for $\mathrm{S}_{1} g=-1 \mathrm{f}=2$
The equation for a circle has the generic form: ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. The coordinates of the circle's center and radius are found using this general form, where g, f, and c are constants. The general form of the equation of a circle makes it difficult to identify any significant properties about any specific circle, in contrast to the standard form, which is simpler to comprehend. So, to quickly change from the generic form to the standard form, we will use the completing the square formula.
$\therefore$ Equation of circle $\mathrm{S}_{1}$ is $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}+4 \mathrm{y}+\mathrm{c}=0$ as it passes through $(4,-2)$
$\therefore x^{2}+y^{2}-2 x+4 y+c=0$
$\Rightarrow 4^{2}+(-2)^{2}-2 \times 4+4 \times-2+c=0$
$\Rightarrow 16+4-8-8+c=0$
$\Rightarrow \mathrm{c}=-4$
Option ‘C’ is correct
Note: If two objects in geometry have a common center, they are said to be concentric. Due to their shared center, regular polygons, regular polyhedra, and circles are all concentric. Two circles that are concentric in Euclidean geometry always have different radii but the same center.
Circles having the same or a shared center are known as concentrators. In other terms, circles are said to be concentric if two or more of them share the same center.
Recently Updated Pages
Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Algebra Made Easy: Step-by-Step Guide for Students

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

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 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

How to Convert a Galvanometer into an Ammeter or Voltmeter

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

