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Question: The line \(2x - y + 6 = 0\) meets the circle \({x^2} + {y^2} - 2y - 9 = 0\) at A and B. Find the equ...

The line 2x−y+6=02x - y + 6 = 0 meets the circle x2+y2−2y−9=0{x^2} + {y^2} - 2y - 9 = 0 at A and B. Find the equation of the circle on AB as diameter.

Explanation

Solution

Since from the given two equation we substitute one equation into another equation and then we can able to values of x,yx,y after that we will apply in into the circle of diameter formula (x−x1)(x−x2)+(y−y1)(y−y2)=0(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0 and hence we solve this to get the equation of the circle on AB as diameter.

Complete step by step solution:
Let from the given we have two equations as 2x−y+6=02x - y + 6 = 0 and x2+y2−2y−9=0{x^2} + {y^2} - 2y - 9 = 0
Convert the equation one values into one side we have 2x−y+6=02x - y + 6 = 0
⇒y=2x+6\Rightarrow y = 2x + 6
Now substitute the converted values into the two equations we get x2+y2−2y−9=0{x^2} + {y^2} - 2y - 9 = 0
⇒x2+(2x+6)2−2(2x+6)−9=0\Rightarrow {x^2} + {(2x + 6)^2} - 2(2x + 6) - 9 = 0
Thus, further solving we have
x2+4x2+36+24x−4x−12−9=0{x^2} + 4{x^2} + 36 + 24x - 4x - 12 - 9 = 0
Using the addition operation on the variables we get
5x2+20x+15=05{x^2} + 20x + 15 = 0
Taking the common numbers as it is multiplied by the number 55 and thus we get
5(x2+4x+3)=05({x^2} + 4x + 3) = 0
Turning the number into right-hand side we get the number as zero because ab=0⇒b=0ab = 0 \Rightarrow b = 0
Hence, we get 5(x2+4x+3)=05({x^2} + 4x + 3) = 0
⇒x2+4x+3=0\Rightarrow {x^2} + 4x + 3 = 0
Converting the term 4x4x into 3x+x3x + x then we get x2+4x+3=0{x^2} + 4x + 3 = 0
⇒x2+3x+x+3=0\Rightarrow {x^2} + 3x + x + 3 = 0
Again, taking the common values out we get x2+3x+x+3=0{x^2} + 3x + x + 3 = 0
⇒x(x+3)+(x+3)=0\Rightarrow x(x + 3) + (x + 3) = 0
Further solving we get x(x+3)+(x+3)=0⇒(x+3)(x+1)=0x(x + 3) + (x + 3) = 0 \Rightarrow (x + 3)(x + 1) = 0
Thus, solving we have (x+3)(x+1)=0(x + 3)(x + 1) = 0
⇒x=−1,−3\Rightarrow x = - 1, - 3
Substituting the values into the first equation y=2x+6y = 2x + 6 then we get
y=4,0y = 4,0
Hence applying these points in the diametrically ends points we get
(x−x1)(x−x2)+(y−y1)(y−y2)=0⇒(x+1)(x+3)+(y−4)(y−0)=0(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0 \Rightarrow (x + 1)(x + 3) + (y - 4)(y - 0) = 0
Further solving the equation, we have x2+4x+3+y2−4y=0{x^2} + 4x + 3 + {y^2} - 4y = 0
Hence simplifying we get x2+y2+4x−4y+3=0{x^2} + {y^2} + 4x - 4y + 3 = 0 which is the required diameter of the given circle equation.

Note:
We first assume the given equation into one value so that we can make use of them to find the values of x,yx,y after finding one of the variables we just substitute it into the main equation to get another equation. Finally, we just apply all the values in the circle diameter equation and which is the required answer.