Equation of a circle given one point and two lines

Question

Find the equation of the circle that pass through $(2,3)$ and are tangent to both the lines $3x - 4y = -1$ and $4x + 3y = 7$.

The center must be equidistant from the point and the lines. — Tony Piccolo, Feb 02 '14 at 07:55

score 1 · Answer 1 · answered Feb 02 '14 at 15:37

Using the angle bisector equation, we obtain that the center of the required circle has to lie on the line $\frac{3x-4y+1}{5} = \pm \frac{4x+3y-7}{5}$. Draw a figure and you can see that we need to consider the equation with minus on RHS. Hence, the required angle bisector $L : 7x-y-6=0$.Or, the center is of the form $(x, 7x-6)$.

We know that the center has to be equidistant from the point $(2,3)$ and both the lines. This condition gives $(x-2)^2 + (7x-6-3)^2 = \frac{(3x - 4(7x- 6) + 1)^2}{25}$ which on simplification gives $x = 2 \text{ or } 6/5$. You can easily verify from the diagram that the required point is $(2,8)$. And the radius $r = 5$.

dato datuashvili · Answer 2 · 2014-02-02T11:20:38.450

0

i would solve this problem right this

first of equation of this circle is

$(x-2)^2+(y-3)^2=r^2$

now it is tangent of lines

$3*x-4*y+1=0$

$4*x+3*y-7=0$

that means that distance from center to point intersection of circle and lines must be equal to each other,in other word ,distance from center to $1$ line and distance from center to another line must be equal to each other,

distance formula from point $(x_0,y_0)$ to line $A*x+b*y+c=0$

is

$(A*x_0+b*y_0+c)/\sqrt{A^2+b^2}$

could you continue from this?also please pay attention that this distance is the same as radius

edited Feb 02 '14 at 11:20

answered Feb 02 '14 at 08:02

dato datuashvili

9,194

1

Yes. Thank you. That's my idea alaso. Thaank you! – Rigoo Feb 02 '14 at 09:21
welcome then,good lucks – dato datuashvili Feb 02 '14 at 09:26
Wait, how to get the point of intersection of line and circle? – Rigoo Feb 02 '14 at 09:36
no you dont need it there,just consider that distance from center of circle to lines are equal to each other and also to $(x-2)^2+(y-3)^2=r^2$,will not have $r^2$ and you will simplify things – dato datuashvili Feb 02 '14 at 09:44
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I thought, (2,3) is not a center but it is only a point on a circle. Isn't it? – Rigoo Feb 02 '14 at 10:25
@Rigoo Here the unknown $(x,y)$ is the center. Moreover the formula for the point-line distance doesn't include the orthogonal projection of the point on the line. – Tony Piccolo Feb 02 '14 at 10:41
Yes. I know but how will I solve it? – Rigoo Feb 02 '14 at 10:45
can you updated which equation you have? – dato datuashvili Feb 02 '14 at 10:49
@Rigoo There are three constraints: $;d(C,t_1)=d(C,P);$,$;d(C,t_2)=d(C,P);$,$;d(C,t_1)=d(C,t_2);$. Coupling two of them is enough. – Tony Piccolo Feb 02 '14 at 11:02

Equation of a circle given one point and two lines

2 Answers2