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  • Find the equation of the circle passing through the points (2,4) and (5,7) and whose centre is on line <img alt="3x-y= 8" src="https://entrancecorner.oncodecogs.com/gif.latex?3x-y%3D%208"

Find the equation of the circle passing through the points (2,4) and (5,7) and whose centre is on line 3x-y= 8.

Option: 1

x^{2}+y^{2}-4x-2y-3= 0


Option: 2

x^{2}+y^{2}+4x+2y-3= 0


Option: 3

x^{2}+y^{2}-4x+2y+3= 0


Option: 4

x^{2}+y^{2}+4x-2y+3= 0


Answers (1)

best_answer

Let the centre of the circle be \left ( a,b \right ) , and let the radius be r . Then the equation of the circle is: \left ( x-a \right )^{2}+\left ( y-b \right )^{2}= r^{2}

We know that the circle passes through the points \left ( 2,4 \right ) and \left ( 5,7 \right ), so we can plug these values into the equation above to get two equations:
\left ( 2-a \right )^{2}+\left ( 4-b \right )^{2}= r^{2}
\left ( 5-a \right )^{2}+\left ( 7-b \right )^{2}= r^{2}

Expanding these equations, we get:
a^{2}-4a+b^{2}-8b+20= r^{2}
a^{2}-10a+b^{2}-14b+74= r^{2}

Subtracting the first equation from the second, we get:
6a-6b+54= 0

Simplifying this equation, we get:
a-b+9= 0

Since \left ( a,b \right )  lies on the line 3x-y= 8,   we can substitute y= 3x-8 into this equation to get:
a-\left ( 3a-8 \right )+9= 0

Simplifying this equation, we get:
a= \frac{17}{4}
Substituting this value of a into the equation a-b+9= 0, we get:
b= \frac{49}{4}
Now we can substitute the values of a and b into one of the equations
\left ( 2-a \right )^{2}+\left ( 4-b \right )^{2}= r^{2}\: or\: \left ( 5-a \right )^{2}+\left ( 7-b \right )^{2}= r^{2} to find r^{2}.  Let's use the first equation:
\left ( 2-\frac{17}{4} \right )^{2}+\left ( 4-\frac{49}{4} \right )^{2}= r^{2}

Simplifying this equation, we get:
r^{2}= \frac{125}{16}

Now we can write the equation of the circle:
\left ( x-\frac{17}{4} \right )^{2}+\left ( y-\frac{49}{4} \right )^{2}= \frac{125}{16}

Expanding and simplifying, we get:
x^{2}+y^{2}-17x-49y+\frac{457}{16}= 0

Therefore, the answer is: d) x^{2}+y^{2}+4x-2y+3= 0


 

 

 

Posted by

jitender.kumar

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