A circle touches two perpendicular lines $2 x - 3 y = 15$ $2x-3y=15$ and $3 x + 2 y = 3$ $3x+2y=3$ at the points $A (6, - 1)$ $A(6,-1)$ , $B (1, 0)$ $B(1,0)$ respectively. Find the equation of the circle?

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A circle touches two perpendicular lines $2 x - 3 y = 15$ $2x-3y=15$ and $3 x + 2 y = 3$ $3x+2y=3$ at the points $A (6, - 1)$ $A(6,-1)$ , $B (1, 0)$ $B(1,0)$ respectively. Find the equation of the circle?

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Answer 1 · 2018-03-31T16:37:56

$x^{2} + y^{2} - 8 x - 4 y + 7 = 0$ $x^2+y^2-8x-4y+7=0$

Explanation:

The tangent to a circle

$x^{2} + y^{2} + 2 a x + 2 b y + c = 0$ $x^2+y^2+ 2ax+2by+c = 0$

at the point $(x_{1}, y_{1})$ $(x_1,y_1)$ is given by

$x x_{1} + y y_{1} + a (x + x_{1}) + b (y + y_{1}) + c = 0$ $x x_1+yy_1+a(x+x_1)+b(y+y_1)+c = 0$

Thus, the tangent at $A = (6, - 1)$ $A = (6,-1)$ is

$6 x - y + a (x + 6) + b (y - 1) + c = 0$ $6x-y+a(x+6)+b(y-1)+c = 0$

or

$(6 + a) x + (b - 1) y + (6 a - b + c) = 0$ $(6+a)x+(b-1)y+(6a-b+c) = 0$

Since this is the line $2 x - 3 y = 15$ $2x-3y = 15$ , we must have

$\frac{6 + a}{2} = \frac{b - 1}{- 3} = \frac{6 a + b - c}{- 15}$ $(6+a)/2 = (b-1)/(-3) = (6a+b-c)/(-15)$

Again, the tangent at $B = (1, 0)$ $B = (1,0)$ is

$1 x + 0 \times y + a (x + 1) + b (y + 0) + c = 0$ $1x+0times y+a(x+1)+b(y+0)+c = 0$

or

$(1 + a) x + b y + (a + c) = 0$ $(1+a)x+by+(a+c) = 0$

Since this is the line $3 x + 2 y = 3$ $3x+2y = 3$ , we must have

$\frac{1 + a}{3} = \frac{b}{2} = \frac{a + c}{- 3}$ $(1+a)/3 = (b)/(2) = (a+c)/(-3)$

Hence $b = \frac{2}{3} (1 + a)$ $b = 2/3(1+a)$ and substituting this in $\frac{6 + a}{2} = \frac{b - 1}{- 3}$ $(6+a)/2 = (b-1)/(-3)$ gives

$\frac{6 + a}{2} = \frac{\frac{2}{3} (1 + a) - 1}{- 3} = - \frac{2}{9} a + \frac{1}{9} \Rightarrow (\frac{1}{2} + \frac{2}{9}) a = \frac{1}{9} - 3$ $(6+a)/2 = (2/3(1+a)-1)/(-3) = -2/9 a+1/9 implies (1/2+2/9)a = 1/9-3$

Thus

$\frac{13}{18} a = - \frac{26}{9} \Rightarrow a = - 4$ $13/18a = -26/9 implies a= -4$

So,

$b = \frac{2}{3} (a + 1) = - 2$ $b = 2/3(a+1) = -2$

and

$\frac{b}{2} = \frac{a + c}{- 3} \Rightarrow - 4 + c = - \frac{3}{2} \times (- 2) = 3 \Rightarrow c = 7$ $(b)/(2) = (a+c)/(-3) implies -4+c = -3/2times (-2) = 3 implies c = 7$

Thus the circle is

$x^{2} + y^{2} - 8 x - 4 y + 7 = 0$ $x^2+y^2-8x-4y+7=0$

Answer 2 · 2018-03-31T16:45:00

${(x - 4)}^{2} + {(y - 2)}^{2} = {(\sqrt{13})}^{2}$ $(x-4)^2+(y-2)^2=(sqrt13)^2$

Explanation:

The following is a graph of:

$2 x - 3 y = 15 [1]$ $color(red)(2x-3y=15)" [1]"$

$3 x + 2 y = 3 [2]$ $color(blue)(3x+2y=3)" [2]"$

$A (6, - 1)$ $color(green)(A(6,-1))$ , and $B (1, 0)$ $color(orange)(B(1,0))$

![www.desmos.com/calculator]( useruploads.socratic.org )

Because the center must be located on lines that are perpendicular to the points of tangency, we can use equations [1] and [2] to write two equations that must intersect at the center.

Set equation [1] equal to an arbitrary constant, $c_{1}$ $c_1$ :

$2 x - 3 y = c_{1} [1.1]$ $2x-3y= c_1" [1.1]"$

Set equation [2] equal to an arbitrary constant $c_{2}$ $c_2$ :

$3 x + 2 y = c_{2} [2.1]$ $3x+2y=c_2" [2.1]"$

Substitute point B into equation [1.1] and solve for $c_{1}$ $c_1$

$2 (1) - 3 (0) = c_{1}$ $2(1)-3(0)= c_1$

$c_{1} = 2$ $c_1 = 2$

Substitute the above into equation [1.1]:

$2 x - 3 y = 2 [1.2]$ $2x-3y= 2" [1.2]"$

Substitute point A into equation [2.1] and solve for $c_{2}$ $c_2$ :

$3 (6) + 2 (- 1) = c_{2}$ $3(6)+2(-1)=c_2$

$c_{2} = 16$ $c_2=16$

Substitute the above into equation [2.1]:

$3 x + 2 y = 16 [2.2]$ $3x+2y=16" [2.2]"$

Add equations [1.2] and [2.2] to the graph:

![www.desmos.com/calculator]( useruploads.socratic.org )

Please understand that the center of the circle must be at the intersection of equations [1.2] and [2.2], therefore, we shall solve them as a system of equations:

$2 x - 3 y = 2 [1.2]$ $2x-3y= 2" [1.2]"$
$3 x + 2 y = 16 [2.2]$ $3x+2y=16" [2.2]"$

$4 x - 6 y = 4 [1.3]$ $4x-6y= 4" [1.3]"$
$9 x + 6 y = 48 [2.2]$ $9x+6y=48" [2.2]"$

$13 x = 52$ $13x=52$

$x = 4$ $x = 4$

$3 (4) + 2 y = 16$ $3(4)+2y=16$

$2 y = 4$ $2y=4$

$y = 2$ $y = 2$

The center of the circle is at the point $C (4, 2)$ $C(4,2)$

The standard Cartesian equation of a circle is:

${(x - h)}^{2} + {(y - k)}^{2} = r^{2} [3]$ $(x-h)^2+(y-k)^2=r^2" [3]"$

where $(x, y)$ $(x,y)$ is any point on the circle, $(h, k)$ $(h,k)$ is the center, and $r$ $r$ is the radius.

Substitute point C into equation [3]:

${(x - 4)}^{2} + {(y - 2)}^{2} = r^{2} [3.1]$ $(x-4)^2+(y-2)^2=r^2" [3.1]"$

To find the radius, we shall substitute point B into equation [3.1]:

${(1 - 4)}^{2} + {(0 - 2)}^{2} = r^{2}$ $(1-4)^2+(0-2)^2=r^2$

$9 + 4 = r^{2}$ $9+4=r^2$

$13 = r^{2}$ $13 = r^2$

$r = \sqrt{13}$ $r = sqrt13$

Substitute the above into equation [3.1]:

${(x - 4)}^{2} + {(y - 2)}^{2} = {(\sqrt{13})}^{2} [3.2]$ $(x-4)^2+(y-2)^2=(sqrt13)^2" [3.2]"$

Add equation [3.2] to the graph:

![www.desmos.com/calculator]( useruploads.socratic.org )

Please observe that equation [3.2] touches [1] and [2] and points A and B respectively, therefore, equation [3.2] is the correct equation of the circle.

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A circle touches two perpendicular lines $2 x - 3 y = 15$ $2x-3y=15$ and $3 x + 2 y = 3$ $3x+2y=3$ at the points $A (6, - 1)$ $A(6,-1)$ , $B (1, 0)$ $B(1,0)$ respectively. Find the equation of the circle?

2 Answers

Explanation:

Explanation:

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A circle touches two perpendicular lines 2x−3y=152x-3y=15 and 3x+2y=33x+2y=3 at the points A(6,−1)A(6,-1), B(1,0)B(1,0) respectively. Find the equation of the circle?

2 Answers

Explanation:

Explanation:

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A circle touches two perpendicular lines $2 x - 3 y = 15$ $2x-3y=15$ and $3 x + 2 y = 3$ $3x+2y=3$ at the points $A (6, - 1)$ $A(6,-1)$ , $B (1, 0)$ $B(1,0)$ respectively. Find the equation of the circle?