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)

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.)