
The equation of the circle with center at (1,-2 ) and passing through the center of the given circle ${{x}^{2}}+{{y}^{2}}-2x+4y+3=0$ is
( a ) ${{x}^{2}}+{{y}^{2}}-2x+4y+3=0$
( b ) ${{x}^{2}}-{{y}^{2}}-2x+4y-3=0$
( c ) ${{x}^{2}}+{{y}^{2}}+2x-4y-3=0$
( d ) ${{x}^{2}}+{{y}^{2}}+2x-4y+3=0$
Answer
163.8k+ views
Hint: To solve this question, we use the standard form of circle and by finding out the values of f,c and g by comparing the equation with the standard form of circle and putting the values in the formula of radius, we get our desirable equation.
Complete Step by step solution:
We have given the circle with centre at (1,-2) which passess through the centre of the given circle ${{x}^{2}}+{{y}^{2}}+2y-3=0$--------- (1)
We have to find the equation of the circle.
We know the standard form of circle = ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$--------- (2)
Compare the equation (1) with equation (2), we get
2g = 0 and 2f = 2
Hence,
g = 0 and f = 1
And we get c = -3
We know center = (-g,-f) = ( 0, -1 )
Now we find the equation of circle that has a center of (1,-2) and passess through (0,-1)
We know that the equation of circle is given by
${{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$
(a,b) is the center of circle is (1,-2) and r is the radius of circle.
That is equation of circle = $1{{(0-1)}^{2}}+{{(-1+2)}^{2}}={{r}^{2}}$
That is 1+1 = ${{r}^{2}}$
This means ${{r}^{2}}$ = 2
So, the required equation of circle = ${{(x-1)}^{2}}+{{(y+2)}^{2}}=2$
By expanding the above equation, we get
${{x}^{2}}-2x+1+{{y}^{2}}+4y+4=2$
That is ${{x}^{2}}+{{y}^{2}}-2x+4y+3=0$
Thus, Option ( A ) is correct.
Note: Students made mistakes while finding the values and putting the right equations. We must have a lot of practice to solve these types of questions.
Complete Step by step solution:
We have given the circle with centre at (1,-2) which passess through the centre of the given circle ${{x}^{2}}+{{y}^{2}}+2y-3=0$--------- (1)
We have to find the equation of the circle.
We know the standard form of circle = ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$--------- (2)
Compare the equation (1) with equation (2), we get
2g = 0 and 2f = 2
Hence,
g = 0 and f = 1
And we get c = -3
We know center = (-g,-f) = ( 0, -1 )
Now we find the equation of circle that has a center of (1,-2) and passess through (0,-1)
We know that the equation of circle is given by
${{(x-a)}^{2}}+{{(y-b)}^{2}}={{r}^{2}}$
(a,b) is the center of circle is (1,-2) and r is the radius of circle.
That is equation of circle = $1{{(0-1)}^{2}}+{{(-1+2)}^{2}}={{r}^{2}}$
That is 1+1 = ${{r}^{2}}$
This means ${{r}^{2}}$ = 2
So, the required equation of circle = ${{(x-1)}^{2}}+{{(y+2)}^{2}}=2$
By expanding the above equation, we get
${{x}^{2}}-2x+1+{{y}^{2}}+4y+4=2$
That is ${{x}^{2}}+{{y}^{2}}-2x+4y+3=0$
Thus, Option ( A ) is correct.
Note: Students made mistakes while finding the values and putting the right equations. We must have a lot of practice to solve these types of questions.
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
