Question

The equation of the circle that can be inscribed in the square formed by $x^2-8x+12=0$ and $y^2-14y+45=0$ is ?

• A. $x^2-8x-14y+61=0$
• B. $x^2-8x-14y+71=0$
• C. $x^2-4x-7y+61=0$
• D. $x^2-4x-7y+71=0$
• E. $x^2+8x+14y -61=0$
• F. None

Answer 1

Answer: (F)

None

Answer 2

Given data

The equation of the circle that can be inscribed in the square formed by $x^2-8x+12=0$ and $y^2-14y+45=0$ can be determine as follows:

The center is given by the intersection of the diagonals i.e the mid -point of a diagonal.

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

$⇒(x-6)(x-2)=0$

$⇒x=6, x=2$

and for

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

$⇒(y-9)(y-5)=0$

$⇒y=5, y=9$

Therefore the the points $(2,5),(2,9),(6,5,),(6,9)$ form squire.

Hence Mid point of diagonal is $=\dfrac{2+6}{2}, \dfrac{5+9}{2}$ $=(4,7)$⇢ This is the center of the circle

Then, mid point of the line is $=\dfrac{2+2}{2}, \dfrac{5+9}{2} =(2,7)$

Now we can find the $d =√((4-2)^{2}+(2-2)^{2})=2$ $units.$

then the equation of the circle is $; (x-4)^{2}+(y-3)^{2}=2^{2}$

$⇒x^2+y^2-8x-6y+25-4=0$

$⇒x^2+y^2-8x-6y+21=0$

is the desired equation for the circle. (Ans.)