I have to find the equation of the circle which passes through the points $A(-2,2)$ and $B(5,-5)$ and has the line $3x-4y=35$ as a tangent at the point $B(5,-5)$.
I tried forming the tangent in a third equation using $(-g,-f)$. With the two points on the circle I used them to form an equation in the form $x^2+y^2+2gx+2fy+c$. I don't know how to go further.
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The centre of such a circle must lie on the perpendicular bisector of $AB$, i.e. on the line $y=x-3$, and on the perpendicular to $3x-4y=35$ through $B$, i.e. on the line $4x+3y=5$. It follows that the centre of the circle lies at $\left(\frac{14}{13},-\frac{25}{13}\right)$. Can you finish from there?
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Hint. The center of the circle is on the perpendicular to the tangent line. Find the equation to the perpendicular to the tangent line and you can then express the y-coordinate of the center in terms of the x-coordinate (or vice versa). Then you can express the circle in terms of just the x-coordinates and point (-2, 2) and in terms of the x-coordinate and point (5, -5). This is two quadratic equations with one variable and should be straightforward to solve for the x-coordinate. The y-coordinate and radius follow easily.
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The equation of the line perpendicular to the given tangent at $B (5,-5)$ can be found to be $4x + 3y = 5$. The centre of the circle must lie on this line. The centre must also lie on the perpendicular bisector of the line joining $A$ and $B$. The equation of this perpendicular can be found by using the fact that it passes through the mid point of $AB$ and perpendicular to $AB$. The equation of this line will be $x - y = 3$. Solving the linear equation gives the centre as $(2,-1)$. The radius is $\sqrt{4^2+(-3)^2} = 5$.
So the equation of circle is $(x-2)^2 +(y+1)^2 = 25$ which can be expressed in standard form as $x^2 + y^2 - 4x + 2y -20 = 0$. This is the answer.
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Slope of AB $m_1 = -7/7 = -1$ , slope of tangent at $ B= m_2=3/4$
tan ( angle between chord AB and tangent at $B $
$$= \dfrac{3/4-(-1)}{1+(-1)(3/4)} = 7 = \tan \gamma;\, \cos \gamma =\frac{1}{\sqrt{50}} $$
We can use a property of circle: Angle in the alternate segment equals that between tangent at B and chord AB. Let a point on circle be C(x,y).
Equate the cosine of the angle of above given known vectors to $\cos \gamma$ inside segment angles between vectors AC and BC ( vector dot product/product of vector absolute values).
$$ AC: (x+2)i + (y-2) j ;\,\, BC (x-5) i + (y+5) j ;\,$$
Hint. Find the equation of every circles passing through $A$ and $B$. Then find the one whose intersection with the line yields only one solution. Since this is a second degree equation, the discriminant needs to be 0.