
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
164.1k+ 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.
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