Question

How do you write the standard from of the equation of the circle given center is at (-1,2) and the point (3,4) lies on the circle?

Answer 1

`(x+1)^2+(y-2)^2=20`

Explanation:

`"the standard form of the equation of a circle is"`

`•color(white)(x)(x-a)^2+(y-b)^2=r^2`

`"where "(a,b)" are the coordinates of the centre and r"`
`"the radius"`

`"here "(a,b)=(-1,2)`

`"the radius is the distance from the centre to a point"`
`"on the circumference"`

`"to calculate r use the "color(blue)"distance formula"`

`•color(white)(x)r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

`"let "(x_1,y_1)=(-1,2)" and "(x_2,y_2)=(3,4)`

`rArrr=sqrt((3+1)^2+(4-2)^2)=sqrt(16+4)=sqrt20`

`rArr(x-(-1))^2+(y-2)^2=(sqrt20)^2`

`rArr(x+1)^2+(y-2)^2=20larrcolor(red)"equation of circle"`