Find the equation of the tangent to the circle $x^{2} + y^{2} - 30 x + 6 y + 109 = 0$ at (4, -1)

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Solution

Equation of tangent is
$4 x + (- y) - 30 (\frac{x + 4}{2}) + 6 (\frac{y + (- 1)}{2}) + 109 = 0$
or $4 x - y - 15 x - 60 + 37 - 3 + 109 = 0 o r - 11 x + 2 y + 46 = 0$
or $11 x - 2 y - 46 = 0$
Hence the required equation of the tangent is $11 x - 2 y - 46 = 0$

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If the point of contact of the circle $x^{2} + y^{2} - 30 x + 6 y + 109 = 0$ with the tangent 11x - 2y - 46 = 0 is (a, b), find a + b

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Q. The equation of the tangent to the circle

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

(2, 8)

Q. Find the equation to the pair of transverse common tangents to the circles

x^{2} + y^{2} - 4 x - 10 y + 28 = 0 a n d x^{2} + y^{2} + 4 x - 6 y + 4 = 0

The angle between a pair of tangents drawn from a point P to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 s i n^{2} α + 13 c o s^{2} α = 0 i s 2 α .$ The equation of the locus of the point P is

Q. I : The equations to the direct common tangents to the circles

x^{2} + y^{2} + 6 x + 4 y + 4 = 0

x^{2} + y^{2} - 2 x = 0

are

y - 1 = 0

4 x - 3 y - 9 = 0

II : The equations to the transverse common tangents to the circles

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x^{2} + y^{2} + 4 x - 6 y + 4 = 0

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x - 1 = 0, 3 x + 4 y - 21 = 0

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Find the equation of the tangent to the circle x2+y2−30x+6y+109=0 at (4, -1)

Equation of tangent is 4x+(−y)−30(x+42)+6(y+(−1)2)+109=0or 4x−y−15x−60+37−3+109=0or−11x+2y+46=0or 11x−2y−46=0Hence the required equation of the tangent is 11x−2y−46=0

Find the equation of the tangent to the circle $x^{2} + y^{2} - 30 x + 6 y + 109 = 0$ at (4, -1)

Equation of tangent is
$4 x + (- y) - 30 (\frac{x + 4}{2}) + 6 (\frac{y + (- 1)}{2}) + 109 = 0$
or $4 x - y - 15 x - 60 + 37 - 3 + 109 = 0 o r - 11 x + 2 y + 46 = 0$
or $11 x - 2 y - 46 = 0$
Hence the required equation of the tangent is $11 x - 2 y - 46 = 0$