please see attachment for qustions:
1. The axes touch the circle in the points P(r,0) and Q(0,r). The center of the circle must be situated in the 3rd quadrant and on the line y = x.
2. The circle has the equation
$\displaystyle (x-a)^2+(y-b)^2=r^2$
According to 1) the coordinates of the center must be equal C(-r , -r):
3. Plug in the coordinates of P, Q, C and the coordinmates of the given point:
$\displaystyle (-2+r)^2+(-1+r)^2=r^2$
Solve for r.
4. You'll get 2 circles which satisfy the given conditions.
The axes are tangents to the circle. The point of intersection of the two tangents and the center of the circle determine the angle bisector between the tangents. The angle bisector of the coordinate axes is the straight line defined by y = x (or y = -x respectively)
I've had a lot of trouble to read your attachment. The second question was unreadable for me. Sorry.and wat about my second question???is it too easy???
Yes. Have a look at the attachment. The point (-2, -1) is drawn. And you can show that the distance between (-2, -1) and (-5, -5) (or (-2, -1) and (-1, -1) ) is as long as the corresponding radius.
Is this the text of the 2nd question?:
2)Find the area of an equilateral triangle inscribed in the circle $\displaystyle x^{2}+y^{2}+2gx+2fy+c=0$
Assuming the points to be $\displaystyle A(x_{1},y_{1}),\ B(x_{2},y_{2}),\ C(x_{3},y_{3})$ the area of the triangle is:
$\displaystyle ar(ABC)=\begin{array}{ccc}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}$
and there is a similar form of the equation of a circle but i donot know how to combine them for a solution.
1. Transform the equation of the circle such that you can determine the coordinates of the center and the length of the radius:
$\displaystyle x^2+2gx+y^2+2fy=-c~\implies~(x+g)^2+(y+f)^2=f^2+g^2-c$
Thus $\displaystyle r^2=f^2+g^2-c$
2. The side of an equilateral triangle which is inscribed in a circle has the length $\displaystyle s = r \cdot \sqrt{3}$
The area of such a triangle is $\displaystyle A = \frac12 \cdot \underbrace{r \cdot \sqrt{3}}_{\text{length of base}} \cdot \underbrace{\frac32 \cdot r}_{\text{length of height}} = \frac34 \cdot r^2 \cdot \sqrt{3}$
3. Plug in the term for rē from 1. and you're done.
In this case simple Pythagorean Theorem will do:
1. See attachment. The small right triangle (coloured in blue) is a 30°-60°-right triangle.
2. Therefore the blue line segment must have the length $\displaystyle \tfrac12 r$
3. The 2nd leg of the right angle has the length:
$\displaystyle \sqrt{r^2-\left(\frac12 r\right)^2} = \frac12 r\cdot \sqrt{3}$
4. This 2nd leg is as long as a half side of the equilateral triangle:
$\displaystyle \frac12 s = \frac12 r\cdot \sqrt{3}~\implies~\boxed{s = r \cdot \sqrt{3}}$