Question

the centre of the circle 3x^(2)+3y^(2)-12x-15y+6=0 is ​

Answer 1

The centre for the given circle is $(6,\frac{15}{2})$

Step-by-step explanation:

Given that the equation of circle is $3x^2+3y^2-12x-15y+6=0\hfill (1)$

To find the centre of the given circle :

The general equation for the circle is $ax^2+by^2+2gx+2fy+c=0\hfill (2)$

The formula for centre of circle is (-g,-f)

Now comparing the equations (1) and (2) we get

• Equating the coeffients of x  we have
• 2g=-12
• $g=-\frac{12}{2}$

Therefore g=-6

• Equating the coefficients of y we have
• 2f=-15

Therefore $f=-\frac{15}{2}$

• Now substitute the values of g and f in the formula for centre of circle (-g,-f)
• $((-6),-(-\frac{15}{2}))$
• $(6,\frac{15}{2})$

Therefore the centre for the given circle is $(6,\frac{15}{2})$