
The line $2x - y + 6 = 0$ meets the circle ${x^2} + {y^2} - 2y - 9 = 0$ at A and B. Find the equation of the circle on AB as diameter.
Answer
446.4k+ views
Hint: Since from the given two equation we substitute one equation into another equation and then we can able to values of $x,y$ after that we will apply in into the circle of diameter formula $(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$ 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 $2x - y + 6 = 0$ and ${x^2} + {y^2} - 2y - 9 = 0$
Convert the equation one values into one side we have $2x - y + 6 = 0$
$ \Rightarrow y = 2x + 6$
Now substitute the converted values into the two equations we get ${x^2} + {y^2} - 2y - 9 = 0$
$\Rightarrow {x^2} + {(2x + 6)^2} - 2(2x + 6) - 9 = 0$
Thus, further solving we have
${x^2} + 4{x^2} + 36 + 24x - 4x - 12 - 9 = 0$
Using the addition operation on the variables we get
$5{x^2} + 20x + 15 = 0$
Taking the common numbers as it is multiplied by the number $5$ and thus we get
$5({x^2} + 4x + 3) = 0$
Turning the number into right-hand side we get the number as zero because $ab = 0 \Rightarrow b = 0$
Hence, we get $5({x^2} + 4x + 3) = 0$
$ \Rightarrow {x^2} + 4x + 3 = 0$
Converting the term $4x$ into $3x + x$ then we get ${x^2} + 4x + 3 = 0$
$\Rightarrow {x^2} + 3x + x + 3 = 0$
Again, taking the common values out we get ${x^2} + 3x + x + 3 = 0$
$ \Rightarrow x(x + 3) + (x + 3) = 0$
Further solving we get $x(x + 3) + (x + 3) = 0 \Rightarrow (x + 3)(x + 1) = 0$
Thus, solving we have $(x + 3)(x + 1) = 0$
$ \Rightarrow x = - 1, - 3$
Substituting the values into the first equation $y = 2x + 6$ then we get
$y = 4,0$
Hence applying these points in the diametrically ends points we get
$(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0 \Rightarrow (x + 1)(x + 3) + (y - 4)(y - 0) = 0$
Further solving the equation, we have ${x^2} + 4x + 3 + {y^2} - 4y = 0$
Hence simplifying we get ${x^2} + {y^2} + 4x - 4y + 3 = 0$ 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 $x,y$ 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.
Complete step by step solution:
Let from the given we have two equations as $2x - y + 6 = 0$ and ${x^2} + {y^2} - 2y - 9 = 0$
Convert the equation one values into one side we have $2x - y + 6 = 0$
$ \Rightarrow y = 2x + 6$
Now substitute the converted values into the two equations we get ${x^2} + {y^2} - 2y - 9 = 0$
$\Rightarrow {x^2} + {(2x + 6)^2} - 2(2x + 6) - 9 = 0$
Thus, further solving we have
${x^2} + 4{x^2} + 36 + 24x - 4x - 12 - 9 = 0$
Using the addition operation on the variables we get
$5{x^2} + 20x + 15 = 0$
Taking the common numbers as it is multiplied by the number $5$ and thus we get
$5({x^2} + 4x + 3) = 0$
Turning the number into right-hand side we get the number as zero because $ab = 0 \Rightarrow b = 0$
Hence, we get $5({x^2} + 4x + 3) = 0$
$ \Rightarrow {x^2} + 4x + 3 = 0$
Converting the term $4x$ into $3x + x$ then we get ${x^2} + 4x + 3 = 0$
$\Rightarrow {x^2} + 3x + x + 3 = 0$
Again, taking the common values out we get ${x^2} + 3x + x + 3 = 0$
$ \Rightarrow x(x + 3) + (x + 3) = 0$
Further solving we get $x(x + 3) + (x + 3) = 0 \Rightarrow (x + 3)(x + 1) = 0$
Thus, solving we have $(x + 3)(x + 1) = 0$
$ \Rightarrow x = - 1, - 3$
Substituting the values into the first equation $y = 2x + 6$ then we get
$y = 4,0$
Hence applying these points in the diametrically ends points we get
$(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0 \Rightarrow (x + 1)(x + 3) + (y - 4)(y - 0) = 0$
Further solving the equation, we have ${x^2} + 4x + 3 + {y^2} - 4y = 0$
Hence simplifying we get ${x^2} + {y^2} + 4x - 4y + 3 = 0$ 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 $x,y$ 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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

