Let P (x, y) be any point on the perpendicular bisector of AB.
Then,            PA =PB
\(\Rightarrow \ \sqrt { { \left( x-3 \right) }^{ 2 }+{ \left( y-6 \right) }^{ 2 } } =\sqrt { { \left( x+3 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 } } \)    [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
On squaring both sides, we get  \({ \left( x-3 \right) }^{ 2 }+{ \left( y-6 \right) }^{ 2 }={ \left( x+3 \right) }^{ 2 }+{ \left( y-4 \right) }^{ 2 }\)
\(\Rightarrow \)  x²-6x+9+y²-12y+36 = x²⁺6x+9+y² -8y+16
\(\Rightarrow \)12x+4y-20=0
\(\Rightarrow\) 3x+y-5 = 0     [dividing by 4]   ......(i)

Hence, the equation of the perpendicular bisector of AB is 3x + Y - 5 = O. 
(i) We know that y-coordinate of every point on X-axis is zero.
On putting y = 0 in Eq. (i), we get
3x - 5 = 0  \(\Rightarrow\)  \(x=\frac { 5 }{ 3 } \)
Thus, the perpendicular bisector of AB cuts X-axis at \(\left( \frac { 5 }{ 3 } ,0 \right) \)
(ii) The coordinates of any point on Y-axis are of the form (0, y).
On putting x = 0 in Eq. (i), we get
y-5=0 \(\Rightarrow \)  y = 5
Thus, the perpendicular bisector of AB intersects Y-axis at (0, 5).

We know that the circumcentre of a triangle is equidistant from the vertices of a triangle.Let A (8,6), B (8, - 2) and C (2, - 2) be the vertices of the given triangle
and P(x, Y) be the circumcentre of this triangle.

Then,        PA = PB = PC      [radii of circle]
\(\Rightarrow \) PA² = PB² =PC²
Now, PA² = PB²
\(\Rightarrow \) (x-8)²+(y-6)² =  (x-8)²+(y+2)²    [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
\(\Rightarrow \) x²-16x+64+y²-12y+36 = x²-16x+64+y²+4y+4
\(\Rightarrow \)-12y-4y=4-36    \(\Rightarrow \)  16y=32
\(\therefore \) y=2   and    PB² =PC²
\(\Rightarrow \)(x-8)²+(y+2)² =  (x-2)²+(y+2)²   
\(\Rightarrow \)x²-16x+64+y²+4y+4 = x²-4x+4+y²+4y+4
\(\Rightarrow \) -16x+4x=8-68  
\(\Rightarrow \)12x=60
\(\therefore \)   x=5
So, the coordinates of the circumcenter Pare (5, 2).
Also, circum radius, PA = PB = PC
=\(\sqrt { { \left( 5-8 \right) }^{ 2 }+{ \left( 2-6 \right) }^{ 2 } } =\sqrt { { \left( -3 \right) }^{ 2 }+{ \left( -4 \right) }^{ 2 } } \)
=\(\sqrt { 9+16 } =\sqrt { 25 } =5\) units
Hence, circumradius of the circle is 5 units.

Here, a cone and spherical lead shots are given. Since, lead shots are dropped into the vessel, so the water which flows out from the vessel is equal to the volume of lead shots.
Given, height of the vessel, h = 8 cm and radius of the vessel, r = 5 cm.
\(\therefore\) Volume of water filled in a vessel = Volume of cone
\(\begin{aligned} & =\frac{1}{3} \times \pi \times r^2 \times h \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{4400}{21} \mathrm{~cm}^3 \end{aligned}\)
Also, radius of a spherical lead shot = 0.5 cm
\(\therefore\) Volume of one spherical lead shot \(=\frac{4}{3} \times \pi \times r^3\)
\(=\frac{4}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{11}{21} \mathrm{~cm}^3\)
Let n be the number of lead shots dropped in the vessel, then according to the question,
n \(\times\) Volume of one spherical lead shot
= \(\frac{1}{4}\)\(\times\)Volume of water filled in a vessel
\(\begin{aligned} \Rightarrow & & n \times \frac{11}{21} & =\frac{4400}{21} \times \frac{1}{4} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & n \times 11 & =1100 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & n & =100 \end{aligned}\)
Hence, required number of lead shots is 100.

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