10.  Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.

Answers (1)
P Pankaj Sanodiya

As we know, 

The equation of the circle with centre ( h, k) and radius r is given by ;

(x-h)^2+(y-k)^2=r^2

Given Here,

Condition 1: the circle passes through  points (4,1) and (6,5)

(4-h)^2+(1-k)^2=r^2

(6-h)^2+(5-k)^2=r^2

Here, 

(4-h)^2+(1-k)^2=(6-h)^2+(5-k)^2

(4-h)^2-(6-h)^2+(1-k)^2-(5-k)^2=0

(-2)(10-2h)+(-4)(6-2k)=0

-20+4h-24+8k=0

4h+8k=44

Now, Condition 2:centre is on the line 4x + y = 16.

4h+k=16

From condition 1 and condition 2 

h=3,\:k=4

Now lets substitute this value of h and k in condition 1 to find out r 

(4-3)^2+(1-4)^2=r^2

1+9=r^2

r=\sqrt{10}

So now, the Final Equation of the circle is 

(x-3)^2+(y-4)^2=(\sqrt{10})^2

x^2-6x+9+y^2-8y+16=10

x^2+y^2-6x-8y+15=0

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