Question

Find the equation of the circle concentric with the circle ${x^2} + {y^2} - 8x - 12y + 15 = 0$ and passing through the point (5,4).

Answer 1

Hint: In this particular type of question we need to compare the given equation with the general equation of the circle to get the coordinates of the centre. Since the circles are concentric we further need to find the radius and finally find the equation of the required circle.

Complete step-by-step answer:

We know that the general equation of the circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ , with centre ( -g,-f )
On comparing ${x^2} + {y^2} - 8x - 12y + 15 = 0$ with ${x^2} + {y^2} + 2gx + 2fy + c = 0$, we get
2gx = -8x
$\Rightarrow$ g = -4
And 2fy = -12y
$\Rightarrow$ f = -6
Thus centre of the required circle is (4,6)

Distance between (4,6) and (5,4)
$\sqrt {{{\left( {5 - 4} \right)}^2} + {{\left( {4 - 6} \right)}^2}} \\\ = \sqrt {1 + 4} \\\ = \sqrt 5 \\\$
Thus the radius of the required circle = $\sqrt 5$
Hence, equation of the required circle is
${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = radiu{s^2}$ ( since h, k is the centre )
${\left( {x - 4} \right)^2} + {\left( {y - 6} \right)^2} = {\sqrt 5 ^2} \\\ \Rightarrow {x^2} + 16 - 8x + {y^2} + 36 - 12y = 5 \\\ \Rightarrow {x^2} + {y^2} - 8x - 12y + 47 = 0 \\\$

Note: Remember to recall the properties and general equation of a circle and the fact that concentric circles have a common centre while solving this type of question. We can also check for the equation of the circle by putting the values (5,4) on the final equation. Also note that both equations of circle is used while solving the question which are and ${x^2} + {y^2} + 2gx + 2fy + c = 0$and ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = radiu{s^2}$.