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`x^(2) + (y-4)^(2) = 5` `x^(2) + (y + 4)^(2) = 5` `(x-4)^(2) + y^(2) = 5` `(x +4)^(2) + y^(2) = 5`

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Time | Transcript |
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00:00 - 00:59 | student question is equation of circle with radius root 5 unit Centre lies on y axis and passes to the point 23 let solve this problem the first information am going to use it Centre lies on y axis and y axis centre is there will be zero and Y coordinates and it is route 53 can write equation of circle and x minus zero its creditors X square + Y - x square is equals to rs485 to find ke 335 |

01:00 - 01:59 | 1234 plus 3 minus A square is equal to 5 going to being that product size will get a minus b square is equals to 1 and from where will get 3 minus A is equals to plus minus 1 then when I take time to get a is equal to 2 and when I take -9 will get a is equals to a circle X square + Y - 2 x square is equal to 5 x square + Y - 4 x square is equal to 5 and the preparation of circle with 35 all the information for condition given equation thank you a student |

**Standard equation of circle**

**General Equation of Circle When the circle passes through the origin.**

**General Equation of Circle When the circle touch x axis and y axis .**

**General Equation of Circle When the circle touches both the axis.**

**Important points to remember**

**Find the Equation; radius; center of the circle and its x intercept and y intercept ?**

**Theorem:- Prove that the equation `x^2+y^2+2gx+2fy+c=0` always represent a circle whose centre is `(-g;-f)` and radius `sqrt(g^2+f^2-c)`**

**Diameter form of the circle**

**Parametric form of circle**

**Equation of a circle passing through 3 points**