Question

Wolte equation of cinde whose centre is (1, 2) and
5
radius is 5​

Answer 1

EXPLANATION.

Equation of circle whose center (1,2) and radius = 5.

As we know that,

General equation of circle,

⇒ x² + y² + 2gx + 2fy + c = 0.

⇒ (x - h)² + (y - k)² = r².

Put the value in equation, we get.

⇒ (x - 1)² + (y - 2)² = (5)².

⇒ x² + 1 - 2x + y² + 4 - 4y = 25.

⇒ x² + y² - 2x - 4y - 20 = 0.

                                                                                         

MORE INFORMATION.

The parametric equations of a circle.

(1) = The Parametric equation of a circle x² + y² = r² are x = r cos∅, y = r sin∅.

(2) = The parametric equation of the Circle ⇒ (x - h)² + (y - k)² r². are x = h + r cos∅, y = k + r sin∅.

(3) = Parametric equation of the circle x² + y² + 2gx + 2fy + c = 0. are x = -g + √g² + f² - c Cos∅.,  y = -f + √g² + f² - c Sin∅.

Position of a point with respect to a circle.

The following formulae are also true for parabola and ellipse.

S₁ > 0 ⇒ Point is Outside the circle.

S₁ = 0 ⇒ Point is on the circle.

S₁ < 0 ⇒ Point is inside the circle.

Answer 2

★$\huge\bf{\blue{\underline{Question:-}}}$

• equation of circle whose centre is (1, 2) and radius is 5.

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★$\huge\bf{\red{\underline{Answer:-}}}$

☯ $\large\sf{\pink{x²+y²-2x-4y-20=0}}$

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★$\huge\bf{\green{\underline{Explanation:-}}}$

☯$\large\sf{\green{ General \:equation\: of\: circle,}}$

$\large\sf{x²+y²+2gx+2fy+c=0}$

$\large\sf{(x-h)²+(y-k)²=r²}$

☯$\large\sf{\green{ Then, \:put\: the\: value,}}$

$\large\sf{(x-1)²+(y-2)²=(5)²}$

$\large\sf{x²+1-2x+y²+4-4y=25}$

$\large\sf{x²+y²-2x-4y-20=0}$

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