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}. \\
$
Last updated date: 14th Mar 2023
•
Total views: 303.3k
•
Views today: 6.84k
Answer
303.3k+ 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
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
