
The equation of the circle which passes through the points (2,3) and (4,5 ) and the center lies on the straight line y – 4x + 3 = 0 is
(a) ${{x}^{2}}+{{y}^{2}}+4x-10y+25=0$
(b) ${{x}^{2}}-{{y}^{2}}-4x-10y+25=0$
( c) ${{x}^{2}}+{{y}^{2}}-4x-10y+16=0$
(d) ${{x}^{2}}+{{y}^{2}}-14y+8=0$
Answer
218.7k+ views
Hint: In this question, we have to find the equation of circle which passes through (2,3) and (4,5). To solve this question, we use the standard form of circle and by putting the points from which equation passes, we get the two equations. From there, we find out the values of f,c and g and putting the values in the standard form of circle , we get our answer.
Formula Used:
Standard form of circle = $x^{2}+y^{2}+2gx+2fy+c=0$
Complete Step by step solution:
We have given the points (2,3) and (4,5 ) and the centre lies on straight line y – 4x + 3 = 0
We have to find the equation of the circle.
We know the standard form of circle = ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$--------- (1)
And it is passed through the point $(2,3)$ and $(4,5)$
Therefore, equation of circle is $4+9+4g+6f+c=0$
We get, $4g+6f+c=-13$ --------------------- (2)
And from (4,5), we get $16+25+8g+10f+c=0$
We get, $8g+10f+c=-41$----------------- (3)
Since the centre of the circle (-g, -f) lies on the straight line $y-4x+3=0$
That is $-f+4g+3=0$ -------------------- (4)
That is $g=\dfrac{f-3}{4}$
Now, we put $g=\dfrac{f-3}{4}$ in equation (2), we get
$7f+c=-10$ --------------------- (5)
Now we multiply equation (2) with 2 and subtract it from equation (3), we get
$-2f-c=-15$…………………………………….. (6)
Now by solving equation (5) and (6) with substitution method, we get
$f=-5$ and $c=25$
Now we put the value of f and c in equation (4), we get
$-(-5)+4g+3=0$
$4g+8=0$
$\therefore g=-2$
Now we put all the values in equation (1), we get
${{x}^{2}}+{{y}^{2}}-4x-10y+25=0$
Thus, Option ( B ) is correct.
Note: As in this question, we get the two equations in the form of f and c. Just like as we find the value of x and y by using elimination and substitution method, we also find out the value of f and c by using any of the substitution method.
Formula Used:
Standard form of circle = $x^{2}+y^{2}+2gx+2fy+c=0$
Complete Step by step solution:
We have given the points (2,3) and (4,5 ) and the centre lies on straight line y – 4x + 3 = 0
We have to find the equation of the circle.
We know the standard form of circle = ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$--------- (1)
And it is passed through the point $(2,3)$ and $(4,5)$
Therefore, equation of circle is $4+9+4g+6f+c=0$
We get, $4g+6f+c=-13$ --------------------- (2)
And from (4,5), we get $16+25+8g+10f+c=0$
We get, $8g+10f+c=-41$----------------- (3)
Since the centre of the circle (-g, -f) lies on the straight line $y-4x+3=0$
That is $-f+4g+3=0$ -------------------- (4)
That is $g=\dfrac{f-3}{4}$
Now, we put $g=\dfrac{f-3}{4}$ in equation (2), we get
$7f+c=-10$ --------------------- (5)
Now we multiply equation (2) with 2 and subtract it from equation (3), we get
$-2f-c=-15$…………………………………….. (6)
Now by solving equation (5) and (6) with substitution method, we get
$f=-5$ and $c=25$
Now we put the value of f and c in equation (4), we get
$-(-5)+4g+3=0$
$4g+8=0$
$\therefore g=-2$
Now we put all the values in equation (1), we get
${{x}^{2}}+{{y}^{2}}-4x-10y+25=0$
Thus, Option ( B ) is correct.
Note: As in this question, we get the two equations in the form of f and c. Just like as we find the value of x and y by using elimination and substitution method, we also find out the value of f and c by using any of the substitution method.
Recently Updated Pages
The maximum number of equivalence relations on the-class-11-maths-JEE_Main

A train is going from London to Cambridge stops at class 11 maths JEE_Main

Find the reminder when 798 is divided by 5 class 11 maths JEE_Main

An aeroplane left 50 minutes later than its schedu-class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In an election there are 8 candidates out of which class 11 maths JEE_Main

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

