Give an equation of a circle with center (-2,-3) and the circle is tangent to the line y = -3/4x + 7/4?

Question

I need it to be explained step by step. I am in 9th grade and dont understand how to do it.

Do i have to complete the square? Is there an equation for this problem? How do I do it?

Answer 1

let us find out the radius by finding out the distance between y=-3/4x+7/4 and (-2,-3)
equation of the tangent in the standard form is 4y=-3x+7
=>3x+4y-7=0
perpendicular distance from (-2,-3) to the line 3x+4y-7=0
is=3(-2)+4(-3)-7/rt(3^2+4^2)
=-6-12-7/5
=5 units ignoring the negative sign
the equation of the circle with centre(-2,-3) and radius 5 units
=(x+2)^2+(y+3)^2=5^2

Answer 2

Give an equation of a circle with center (-2,-3) and the circle is tangent to the line y = -3/4x + 7/4

The general equation for a circle with center at (h,k) is
(x-h)^2 + (y-k)^2 = r^2 where r is the radius. So, we have
(x+2)^2 + (y+3)^2 = r^2

Now we need to find r.

The y= -3/4x +7/4 is to be tangent to the circle. Thus the radius will be perpendicular to the line at the point of tangency. So the equation of the radius must be y= 4/3x + b. To find b we use the coordinates of the center of the circle to get -3= 4/3(-2) =b. This give b= -3 +8/3 = -1/3 S o equation of radius is y =4/3x-1/3
When the tangent and the radius intersect we have
4/3x-1/3 = -3/4x +7/4
4/3x +3/4x = 7/4 + 1/3
16/12x + 9/12 x. = 21/12 + 4/12
25/12 x = 25/12
x = 1
y= -3/4(1) + 7/4 = 1
So the point of tangency is (1,1)

The radius the is the distance from the center of the circle to the point of tangency = sqrt((-2-1)^2 + (-3-1)^2)
r = sqrt (9 +16) = sqrt (25) = 5
So, finally the equation of the circle is
(x+2)^2 + (y+3)^2 = 25

Answer 3

the most general form of a two-d circle equation is
(x-a)^2 + (y-b)^2 = r^2 where a,b is the center and r = radius
in the problm you are given a, b , i.e -2 ,-3 so now it becomes
(x+2)^2 + (y+3)^2 = r^2 ... to find r. you must realize that r is the perpendicular distance from the center to the tangentline.. there is a formula to find a distance from a point to a line.. namely

(ax_1 + by_1+c)/(sqrt(a^2 + b^2)) where (x_1,y_1) is the coordinate of the point (which in this case is none other than (-2,-3) and ax+by+c=0 is the line equation.. from the problem that you have... where y = -3/4x +7/4 ..you can change this form of line equation into
3x+4y-7=0 (i hope u know how to do it.. just multiply thru by 4 and bring all the terms into LHS of the equation) so accorfing to what i gave u previously the distance now is

(3x_1 + 4y_1 - 7)/(sqrt (3^2 + 4^2))
= (3(-2) + 4(-3) - 7 )/ sqrt (25)= -25/sqrt(25)
so r^2 is just ((-25)/5)^2 = 25 so now your full equation for the circle is


(x+2)^2 + (y+3)^2 = 25

Answer 4

The equation of a circle is:

(x-x0)² + (y-y0²) = r²

With (x0, y0) the center of the circle and r the radius.
So find the distance from the center point to the line and fill it in in the equation.