Question

find the radius and centre of the circule x^2+y^-6x+8y+9=0​

Answer 1

EXPLANATION.

Equation of circle → x² + y² - 6x + 8y + 9 = 0.

General equation of circle

→ x² + y² + 2gx + 2fy + c = 0.

Centre of the circle = ( -g, -f).

Radius of the circle = √g² + f² - c = 0.

Compare both the equation we get,

→ Centre = ( 3 , -4).

→ radius = √ (3)² + (-4)² - 9 = 0.

→ radius = √ 9 + 16 - 9 = 0.

→ radius = √ 16 = 4.

More information.

$\sf \: case \: = 1 \\ \\ \sf \: \implies \: c_{1} c_{2} > r_{1} \: + \: r_{2} \: \\ \\ \sf \: \implies the \: distance \: between \: centre \: is \: greater \: than \: the \: sum \: of \: radii$

$\sf \: \implies \: case \: = 2 \: \implies \: c_{1} c_{2} \: = r_{1} \: + \: r_{2} \\ \\ \sf \: \implies the \: distance \: between \: the \: centre \: is \: equal \: to \: the \: sum \: of \: radii$

$\sf \: \implies case \: = 3 \implies c_{1} c_{2} < r_{1} \: + \: r_{2} \\ \\ \sf \: \implies \: the \: distance \: between \: the \: centre \: is \: less \: than \: sum \: of \: radii$

$\sf \: \implies \: case \: = 4 \implies \: c_{1}c_{2} \: = | r_{1} \: - \: r_{2} | \\ \\ \sf \: \implies \: the \: distance \: between \: centre \: is \: equal \: to \: the \: difference \: of \: radii$

$\sf \: \implies \: case \: = 5 \implies \: c_{1} c_{2} < | r_{1} \: - \: r_{2} | \\ \\ \sf \: \implies the \: difference \: between \: the \: centre \: is \: less \: than \: the \: difference \: of \: radii$

Answer 2

Correct question= How do you find the center and radius of the circle x2+y2−6x+8y=0?

Solution⬇️

The equation of a circle with radius r and center (h, k) in standard form is (x - h)^2 + (y - k)^2 = r^2.

The equation given is x^2 + y^2 - 6x + 8y + 9 = 0

x^2 + y^2 - 6x + 8y + 9 = 0

=> x^2 - 6x + y^2 + 8y + 9 = 0

=> x^2 - 6x + 9 + y^2 + 8y + 16 = 9 + 16 - 9

=> (x - 3)^2 + (y + 4)^2 = 4^2

This is the equation of a circle with center (3, -4) and radius 4

The standard form of x^2 + y^2 - 6x + 8y + 9 = 0 is (x - 3)^2 + (y + 4)^2 = 4^2