Length of y - intercept made by the circle $5{{\text{x}}^2} + 5{{\text{y}}^2} - 2{\text{x + 6y - 8 = 0 is:}}$
$
{\text{A}}{\text{.}}\dfrac{{19}}{5}. \\
{\text{B}}{\text{.}}\dfrac{{14}}{5}. \\
{\text{C}}{\text{.}}\dfrac{{11}}{5}. \\
{\text{D}}{\text{.}}\dfrac{9}{5}. \\
$
Answer
Verified
505.8k+ views
Hint: In this question, the equation of circle is given. To find the y-intercept made by a circle we will first convert the given circle equation into the standard circle equation and then use the formula for y-intercept to the value of y-intercept.
Complete step-by-step answer:
In the question, it is given that:
Equation of circle is $5{{\text{x}}^2} + 5{{\text{y}}^2} - 2{\text{x + 6y - 8 = 0}}$ .
We have to find the y-intercept made by the given circle.
We know that the standard equation of circle is given by:
${{\text{x}}^2} + {{\text{y}}^2} + 2{\text{gx + 2fy + c = 0}}$. (1)
Y-intercept of this circle is given by $2\sqrt {{{\text{f}}^2} - c} $ .
But the equation of the circle given is not in the standard form. So we will first convert this equation in standard form.
$
\Rightarrow 5{{\text{x}}^2} + 5{{\text{y}}^2} - 2{\text{x + 6y - 8 = 0}} \\
$
On dividing the above equation by 5, we get,
$
\Rightarrow {{\text{x}}^2} + {{\text{y}}^2} - \dfrac{2}{5}{\text{x + }}\dfrac{6}{5}{\text{y - }}\dfrac{8}{5}{\text{ = 0}} \\
$
So the above equation is the standard equation of a circle.
On comparing the above equation with equation 1, we get:
$
2{\text{g = - }}\dfrac{2}{5} \\
\Rightarrow {\text{g = - }}\dfrac{1}{5}. \\
{\text{And}} \\
{\text{2f = }}\dfrac{6}{5}. \\
\Rightarrow {\text{f = }}\dfrac{3}{5},{\text{ and c = - }}\dfrac{8}{5}. \\
$
Y-intercept made by circle =$2\sqrt {{{\text{f}}^2} - c} $.
Putting the value of ‘f’ and ‘c’ in the above formula, we get:
Y-intercept made by circle = $2\sqrt {{{\text{f}}^2} - c} = 2\sqrt {{{\left( {\dfrac{3}{5}} \right)}^2} - \left( { - \dfrac{8}{5}} \right)} = 2\sqrt {\dfrac{9}{{25}} + \dfrac{8}{5}} = 2\sqrt {\dfrac{{49}}{{25}}} = 2 \times \dfrac{7}{5} = \dfrac{{14}}{5}.$
Note: In this type of question, the first important thing is to clearly see the question whether it is asking y-intercept or x-intercept. Then convert the given equation into a standard form of equation. You should remember the formula for finding y-intercept. Compare the transformed given equation with the standard equation to get the value of parameters required for computing the y-intercept.
Complete step-by-step answer:
In the question, it is given that:
Equation of circle is $5{{\text{x}}^2} + 5{{\text{y}}^2} - 2{\text{x + 6y - 8 = 0}}$ .
We have to find the y-intercept made by the given circle.
We know that the standard equation of circle is given by:
${{\text{x}}^2} + {{\text{y}}^2} + 2{\text{gx + 2fy + c = 0}}$. (1)
Y-intercept of this circle is given by $2\sqrt {{{\text{f}}^2} - c} $ .
But the equation of the circle given is not in the standard form. So we will first convert this equation in standard form.
$
\Rightarrow 5{{\text{x}}^2} + 5{{\text{y}}^2} - 2{\text{x + 6y - 8 = 0}} \\
$
On dividing the above equation by 5, we get,
$
\Rightarrow {{\text{x}}^2} + {{\text{y}}^2} - \dfrac{2}{5}{\text{x + }}\dfrac{6}{5}{\text{y - }}\dfrac{8}{5}{\text{ = 0}} \\
$
So the above equation is the standard equation of a circle.
On comparing the above equation with equation 1, we get:
$
2{\text{g = - }}\dfrac{2}{5} \\
\Rightarrow {\text{g = - }}\dfrac{1}{5}. \\
{\text{And}} \\
{\text{2f = }}\dfrac{6}{5}. \\
\Rightarrow {\text{f = }}\dfrac{3}{5},{\text{ and c = - }}\dfrac{8}{5}. \\
$
Y-intercept made by circle =$2\sqrt {{{\text{f}}^2} - c} $.
Putting the value of ‘f’ and ‘c’ in the above formula, we get:
Y-intercept made by circle = $2\sqrt {{{\text{f}}^2} - c} = 2\sqrt {{{\left( {\dfrac{3}{5}} \right)}^2} - \left( { - \dfrac{8}{5}} \right)} = 2\sqrt {\dfrac{9}{{25}} + \dfrac{8}{5}} = 2\sqrt {\dfrac{{49}}{{25}}} = 2 \times \dfrac{7}{5} = \dfrac{{14}}{5}.$
Note: In this type of question, the first important thing is to clearly see the question whether it is asking y-intercept or x-intercept. Then convert the given equation into a standard form of equation. You should remember the formula for finding y-intercept. Compare the transformed given equation with the standard equation to get the value of parameters required for computing the y-intercept.
Recently Updated Pages
A uniform rod of length l and mass m is free to rotate class 10 physics CBSE
Solve the following pairs of linear equations by elimination class 10 maths CBSE
What could be the possible ones digits of the square class 10 maths CBSE
Where was the Great Bath found A Harappa B Mohenjodaro class 10 social science CBSE
PQ is a tangent to a circle with centre O at the point class 10 maths CBSE
The measures of two adjacent sides of a parallelogram class 10 maths CBSE
Trending doubts
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Why is there a time difference of about 5 hours between class 10 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE
Explain the Treaty of Vienna of 1815 class 10 social science CBSE
Write an application to the principal requesting five class 10 english CBSE