Question

What is the center of the circle inscribed in a square formed by lines $$x^2-8x+12=0$$ and $$y^2-14y+45=0$$?
(a). (4, 7)
(b). (7,4)
(c). (9,4)
(d). (4,9)

Answer 1

Hint: Solve the two equations to find the equation of four lines that form the square. Use the concept that the center of the square is the same as the center of the inscribed circle to find the center of the inscribed circle.

Complete step by step answer:
It is given that the square is formed by the following lines:

$$x^2-8x+12=0...........(1)$$

$$y^2-14y+45=0.........(2)$$

Let us solve equation (1) to get the equations of two lines.

$$x^2-8x+12=0$$

We know that – 8x can be split as – 6x and – 2x, then, we have:

$$x^2-6x-2x+12=0$$

Now, we can group the common terms.

 $$x(x-6)-2(x-6)=0$$

$$(x-2)(x-6)=0$$

$$x=2;x=6.........(3)$$

Hence, the equations of the two lines are x = 2 and x = 6.

We now solve equation (2) to find the equation of the other two lines which form the square.

$$y^2-14y+45=0$$

We know that – 14y can be split into – 9y and – 5y, then, we have:

$$y^2-9y-5y+45=0$$

Taking the common terms and simplifying, we get:

$$y(y-9)-5(x-9)=0$$

$$(y-5)(x-9)=0$$

$$y=5;y=9........(4)$$

Hence, the equations of the other two lines are y = 5 and y = 9.

Hence, the lines x =2, x= 6, y = 5 and y = 9 form a square.

We know that the centre of the inscribed circle is the centre of the square itself.

The centre of the square is given as:

$$(x,y)=\left({\displaystyle \frac{2+6}{2}},{\displaystyle \frac{5+9}{2}}\right)$$

$$(x,y)=\left({\displaystyle \frac{8}{2}},{\displaystyle \frac{14}{2}}\right)$$

$$(x,y)=\left(4,7\right)$$

Hence, the centre of the inscribed circle is (4,7).

Note: You can also find the center of the inscribed circle by considering the distance of the center from any of the sides and using the equation of that respective side.