Question

The line 2y-x=12 intersects the circle x2+y2-10x-12y+36=0 at the point A and B. The perpendicular bisector of AB intersects the circle at the point P and Q. What is the exact coordinates of P and Q? What is the exact area of quadrilateral APBQ?

Answer 1

Answer:

Step-by-step explanation:

$\sf{The\,\,given\,\, line\,\,is\,,\,\,L:2y-x-12=0}\\\\\sf{The\,\,given\,\, circle\,\,is\,,\,\,S:x^2+y^2-10x-12y+36=0}$

Now, the line intersects the circle, so,

From line, we have, x = 2y - 12

Put the value of x in the circle,

So,

$\sf{\implies\,\left(2y-12\right)^2+y^2-10\left(2y-12\right)-12y+36=0}$

$\sf{\implies\,4y^2-48y+144+y^2-20y+120-12y+36=0}$

$\sf{\implies\,5y^2-80y+300=0}$

$\sf{\implies\,y^2-16y+60=0}$

$\sf{\implies\,y^2-10y-6y+60=0}$

$\sf{\implies\,y(y-10)-6(y-10)=0}$

$\sf{\implies\,(y-6)(y-10)=0}$

$\sf{\implies\,y=6\,\,\,\,or\,\,\,\,y=10}$

Put this value in the line

So,

$\sf{\implies\,x=0\,\,\,\,or\,\,\,\,x=8}$

So, the coordinates of points A and B are

$\boxed{\tt{A\equiv(0,6)\,\,\,\,and\,\,\,\,B\equiv(8,10)}}$

Now, let the midpoint of AB be M and its coordinates are M ≡ (4, 8)

Slope of the line perpendicular to AB is, $\sf{m=-2}$

So, the equation of the line perpendicular to AB is

$\tt{y-8=(-2)(x-4)}$

$\tt{\implies\,y-8=-2x+8}$

$\tt{\implies\,y=-2x+16}$

Now, this line will cut the circle at two points, which are P and Q

So,

$\sf{x^2+y^2-10x-12y+36=0}$

$\sf{\implies\,x^2+(-2x+16)^2-10x-12(-2x+16)+36=0}$

$\sf{\implies\,x^2+4x^2-64x+256-10x+24x-192+36=0}$

$\sf{\implies\,5x^2-50x+100=0}$

$\sf{\implies\,x^2-10x+20=0}$

$\sf{\implies\,x^2-2\cdot5\cdot\,x+25=5}$

$\sf{\implies\,(x-5)^2=(\sqrt{5})^2}$

$\sf{\implies\,x=5+\sqrt{5}\,\,\,\,or\,\,\,\,x=5-\sqrt{5}}$

Put the values of x in $\sf{y = - 2x + 16}$

So,

$\sf{\implies\,y=6-2\sqrt{5}\,\,\,\,or\,\,\,\,x=6+2\sqrt{5}}$

So, the coordinates of the points P and Q are,

$\boxed{\sf{P\equiv(5+\sqrt{5},6-2\sqrt{5})\,\,\,\,and\,\,\,\,Q\equiv(5-\sqrt{5},6+2\sqrt{5})}}$

For the area of a quadrilateral,

$\tt{A=\dfrac{1}{2}\left|\left|\begin{array}{cc}x_{1}&y_{1}\\x_{2}&y_{2}\\x_{3}&y_{3}\\x_{4}&y_{4}\\x_{1}&y_{1}\end{array}\right|\right|}\\\\\tt{\implies\,A=\dfrac{1}{2}\left|(x_1y_2-y_1x_2)+(x_2y_3-y_2x_3)+(x_3y_4-y_3x_4)+(x_4y_1-y_4x_1)\right|}$

So,

$\tt{A=\dfrac{1}{2}\left|\left|\begin{array}{cc}0&6}\\8&10\\5+\sqrt{5}&6-2\sqrt{5}\\5-\sqrt{5}&6+2\sqrt{5}\\0&6\end{array}\right|\right|}\\\\\tt{\implies\,A=\dfrac{1}{2}\left|(0-48)+(48-16\sqrt{5}-50-10\sqrt{5})+(40+16\sqrt{5}-40+16\sqrt{5})+(30-6\sqrt{5})\right|}$

$\tt{\implies\,A=\dfrac{1}{2}\left|-48-2-26\sqrt{5}+32\sqrt{5}+30-6\sqrt{5}\right|}$

$\tt{\implies\,A=\dfrac{1}{2}\left|-50-32\sqrt{5}+32\sqrt{5}+30\right|}$

$\tt{\implies\,A=\dfrac{1}{2}\left|-50+30\right|}$

$\tt{\implies\,A=\dfrac{1}{2}\left|-20\right|}$

$\tt{\implies\,A=\dfrac{1}{2}\times20}\\\\\tt{\implies\,A=10\,\,sq.\,\,units}$