if the straight line y = x + c is tangent to the circle (x-1

fskilnik@GMATH wrote:If the straight line y = x + c is tangent to the circle (x-1)^2 + (y+2)^2 = 4, what is the maximum possible value for the constant c ?

$$(A) \,\, 1 - \sqrt 2 \,\,\,\,\,\, (B) \,\, 1 + \sqrt 2 \,\,\,\,\,\, (C) \,\, 3 - 2 \sqrt 3 \,\,\,\,\,\, (D) \,\, -3 + 2 \sqrt 2 \,\,\,\,\,\, (E) \,\, -1$$
Source: https://www.gmath.net

\[? = {c_{\max }} = c\,\,\,\,\left( {{\rm{figure}}} \right)\]

The solution I present below is essentially geometric:



\[- 2 - c = 1 - 2\sqrt 2 \,\,\,\,\, \Rightarrow \,\,\,\,? = c = - 3 + 2\sqrt 2\]

I invite other members of this great community to present other approaches.

(I will show an algebraic solution in a couple of days, if nobody offers it before!)

Regards,
Fabio.

fskilnik@GMATH wrote:If the straight line y = x + c is tangent to the circle (x-1)^2 + (y+2)^2 = 4, what is the maximum possible value for the constant c ?

$$(A) \,\, 1 - \sqrt 2 \,\,\,\,\,\, (B) \,\, 1 + \sqrt 2 \,\,\,\,\,\, (C) \,\, 3 - 2 \sqrt 3 \,\,\,\,\,\, (D) \,\, -3 + 2 \sqrt 2 \,\,\,\,\,\, (E) \,\, -1$$

Answer: [spoiler]___(D)__[/spoiler]
Level: 650-700
Source: https://www.gmath.net

The center of the circle is (1, -2). The radius drawn from the center to the tangent point is perpendicular to the tangent, meaning that its slope is the negative inverse of the slope of the tangent line.

The tangent line, Y=X + C has a slope of 1, so the slope of the line from the center to the tangent is -1.

The equation of the line from the center to the tangent point can then be written as Y + 2 =-1(X - 1) or Y = -X -1.

In order to find where this line and the circle intersect, the tangent point, substitute into the circle equation

(X-1)^2 + (1 - X)^2 = 4. Expand this to X^2 -2X - 1 = 0.

Solve for X using the quadratic equation with X = 1+ or -2^1/2.

Plug this X into the circle equation to solve for Y, with Y then = 2^1/2 - 2.

So the point of tangency is (1 + or - 2^1/2, 2^1/2 - 2). This is also on the line Y =X + C. Another point on the line is where X = 0, in which case Y = C.

Using slope = (Y-Y1)/(X-X1) = 1 in this case, we have C-2^1/2+2 = 0-1+ 2^1/2 or 0-1-2^1/2.

Solve for C in both cases: C= 2(2^1/2) -3 or C = -3.

The largest C of these two choices is [spoiler]2(2^1/2) -3, D[/spoiler]

fskilnik@GMATH wrote:If the straight line y = x + c is tangent to the circle (x-1)^2 + (y+2)^2 = 4, what is the maximum possible value for the constant c ?

$$(A) \,\, 1 - \sqrt 2 \,\,\,\,\,\, (B) \,\, 1 + \sqrt 2 \,\,\,\,\,\, (C) \,\, 3 - 2 \sqrt 3 \,\,\,\,\,\, (D) \,\, -3 + 2 \sqrt 2 \,\,\,\,\,\, (E) \,\, -1$$
Source: https://www.gmath.net

\[? = {c_{\max }} = c\]

Algebraic approach:

\[\left\{ \begin{gathered}
\,\left( 1 \right)\,\,{\left( {x - 1} \right)^2} + {\left( {y + 2} \right)^2} = 4 \hfill \\
\,\left( 2 \right)\,\,y = x + c \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 2 \right)\,\,{\text{in}}\,\,\left( 1 \right)} \,\,\,\, \ldots \,\,\,\,\, \Rightarrow \,\,\,\,2{x^2} + 2\left( {c + 1} \right)x + {c^2} + 4c + 1 = 0\]

\[{\text{tangency}}\,\,\, \Rightarrow \,\,\,0 = \Delta = {\left[ {2\left( {c + 1} \right)} \right]^2} - 4\left( 2 \right)\left( {{c^2} + 4c + 1} \right)\,\,\, = \,\,\, \ldots \,\,\, = \,\,\, - 4\left( {{c^2} + 6c + 1} \right)\]

\[{c^2} + 6c + 1 = 0\,\,\,\,\mathop \Rightarrow \limits^{{\text{Bhaskara}}} \,\,\,\left\{ \begin{gathered}
\,{c_1} = - 3 - 2\sqrt 2 \hfill \\
\,{c_2} = - 3 + 2\sqrt 2 \hfill \\
\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {c_2} = - 3 + 2\sqrt 2 \]


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.