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
if the straight line y = x + c is tangent to the circle (x-1
This topic has expert replies
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
- fskilnik@GMATH
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
\[? = {c_{\max }} = c\,\,\,\,\left( {{\rm{figure}}} \right)\]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
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
-
- Master | Next Rank: 500 Posts
- Posts: 415
- Joined: Thu Oct 15, 2009 11:52 am
- Thanked: 27 times
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.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 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
- GMAT Instructor
- Posts: 1449
- Joined: Sat Oct 09, 2010 2:16 pm
- Thanked: 59 times
- Followed by:33 members
\[? = {c_{\max }} = c\]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
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br