Question

T and radius, then grapher nts a circle with a real now uxo on. Ty cus and a the locus the dire SAU b) 3x + 3y - 11x=-91 & x² + y² - Ax = 0 3 x² + y² - 2x By h • 4x + 4y² 1) x² + y² - Ty=0 ) x + y - 2ax +2byc 16x + 2y = 21 chvara d) 36x + 36y? - 36x +48y = -16 gurer Ha h (5, 3) and is concentric with x? par and concentric with the circles? + twice as far from (6,0) as from (0,0). such that the square of the distance points: (0,0), (6,0),(0,-4), and (5,1 -y=27 + 2y = -2 and tangent to the line 5x ent to the circle x² + y2 + 6x + 4y = (b) X = 2

Answer 1

b)

$3x^2+3y^2-11x=-91$

We will arrange the terms first

$3x^2-11x+3y^2=-91$

First, we will divide all the terms by 3 to make the coefficients of x^2 and y^2 equal to 1

$\begin{gathered} \frac{3x^2}{3}-\frac{11}{3}x+\frac{3y^2}{3}=-\frac{91}{3} \\ x^2-\frac{11}{3}x+y^2=-\frac{91}{3} \end{gathered}$

Now, divide the term of x by 2 to find the 2nd term of completing the square

$\because-\frac{\frac{11}{3}}{2}x=-\frac{11}{6}x$

The first term is x and the 2nd term is -11/6

$\because(-\frac{11}{6})^2=\frac{121}{36}$

Then we must add and subtract 121/36

$\therefore x^2-\frac{11}{3}x+\frac{121}{36}-\frac{121}{36}+y^2=-\frac{91}{3}$

Now I will take the first 3 terms and make the bracket power 2 of them

$\because(x^2-\frac{11}{3}x+\frac{121}{36})=(x-\frac{11}{6})^2$$\therefore(x-\frac{11}{6})^2-\frac{121}{36}+y^2=-\frac{91}{3}$

Add both sides by 121/36

$(x-\frac{11}{6})^2+y^2=-\frac{971}{36}$

From this form r ^2 is a negative value then it could not be a real non zero

The equation does not represent a circle with a real non-zero radius