- x^{2} + y^{2} = 25 at (3, -4)
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Solution
Given equation of circle is
\(x^2+y^2=25\)
The radius of the circle is
\(r^2=25\)
Given point is
\((x_1,y_1)=(3,4) \)
Now, the equation of tangent is
\( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
\( y+4 = - \frac{3}{(-4)} (x-3) \)
\(3x-4y=25\)
Next, the equation of normal is (perpendicular to the tangent and passes through origin)
\(4x+3y=0\)
- \( x^2+y^2=4 \) at \( (1,\sqrt{3} ) \)
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Solution
Given equation of circle is
\(x^2+y^2=4\)
The radius of the circle is
\(r^2=4\)
Given point is
\((x_1,y_1)= (1,\sqrt{3} ) \)
Now, the equation of tangent is
\( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
\( y-\sqrt{3} = - \frac{1}{\sqrt{3}} (x-1)\)
\(x+\sqrt{3}y=4\)
Next, the equation of normal is (perpendicular to the tangent and passes through origin)
\(\sqrt{3}x-y=0\)
- \(x^² + y^² =4 \) at \( (2 \cos \theta, 2 \sin \theta) \)
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Solution
Given equation of circle is
\(x^2 + y^2 = 4\)
The radius of the circle is
\(r^2 = 4 \)
Given point is
\((x_1, y_1) = (2 \cos \theta, 2 \sin \theta) \)
Now, the equation of tangent at \((x_1, y_1)\) is
\( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
Substitute \(x_1\) and \(y_1\) into the equation
\( y-2 \sin \theta= - \frac{2 \cos \theta}{2 \sin \theta} (x-2 \cos \theta)\)
\(x \cos \theta + y \sin \theta = 2 \)
Next, the equation of normal is (perpendicular to the tangent and passes through origin)
\(x \sin \theta-y \cos \theta=0\)
- \( x^2+y^2=8 \) at \( (2,2) \)
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Solution
Given equation of circle is
\(x^2 + y^2 = 8\)
The radius of the circle is
\(r^2 = 8 \)
Given point is
\((x_1, y_1) = (2, 2) \)
Now, the equation of tangent at \((x_1, y_1)\) is
\( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
Substitute \(x_1\) and \(y_1\) into the equation
\( y-2 = - \frac{2}{2} (x-2)\)
Simplify
\( x + y = 4 \)
Next, the equation of normal is (perpendicular to the tangent and passes through origin)
\(y-x=0\)
- \( x^2+y^2=36 \) at \( (-6,0) \)
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The radius of the circle is
\(r^2 = 36 \implies r = 6\)
Given point is
\((x_1, y_1) = (-6, 0) \)
Now, the equation of tangent at \((x_1, y_1)\) is
\( y-y_1 = - \frac{x_1}{y_1} (x-x_1)\)
Substitute \(x_1\) and \(y_1\) into the equation
\( y-0 = - \frac{-6}{0} (x-6)\)
Simplifying gives
\(x-6=0 \)
Next, the equation of normal is (perpendicular to the tangent and passes through origin)
\(y=0\)
- \(x^² + y^² + 2 x + 4 y —20 = 0 \) at (3,1)
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Solution
Given equation of circle is
\(x^² + y^² + 2 x + 4 y —20 = 0 \)
The center of the circle is
\((h,k)=(-g,-f)=(-1,-2)\)
The radius of the circle is
\(r^2=g^2+f^2-c=25\)
Given point is
\((x_1,y_1)= (3,1) \)
Now, the equation of tangent is
\( y-y_1 = - \frac{x_1-h}{y_1-k} (x-x_1)\)
\( y-1 = - \frac{3+1}{1+2} (x-3)\)
\(4x+3y=15\)
Next, the equation of normal is (perpendicular to the tangent)
\(3x-4y=k\)
The normal line passes through center \((h,k)=(-g,-f)=(-1,-2)\), so the equation is
\(3x-4y=5\)
- \(x^² + y^² -6x-8y-4 =0\) at (8,6)
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Solution
Given equation of circle is
\(x^2 + y^2 - 6x - 8y - 4 = 0\)
To find the center and radius, we rewrite the equation in standard form
\((x - 3)^2 + (y - 4)^2 = 29\)
The center of the circle is
\((h, k) = (3, 4)\)
The radius of the circle is
\(r^2 = 29 \)
Given point is
\((x_1, y_1) = (8, 6) \)
Now, the equation of tangent is
\( y - y_1 = -\frac{x_1 - h}{y_1 - k}(x - x_1) \)
Substitute \(h = 3\), \(k = 4\), \(x_1 = 8\), and \(y_1 = 6\) into the equation
\( y - 6 = -\frac{8 - 3}{6 - 4}(x - 8) \)
\( y - 6 = -\frac{5}{2}(x - 8) \)
\( 5x + 2y = 52 \)
Next, the equation of normal is (perpendicular to the tangent)
\(2x-5y=k\)
The normal line passes through center \((h,k)=(3, 4)\), so the equation is
\(2x-5y+14=0\)
- \(x^² + y^² -3x+10y-15=0 \) at (4,-11)
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Solution
Given equation of circle is
\(x^2 + y^2 - 3x + 10y - 15 = 0\)
To find the center and radius, we rewrite the equation in standard form
\((x - \frac{3}{2})^2 + (y + 5)^2 = \frac{139}{4} \)
The center of the circle is
\((h, k) = (\frac{3}{2}, -5)\)
The radius of the circle is
\(r^2 = \frac{139}{4} \)
Given point is
\((x_1, y_1) = (4, -11) \)
Now, the equation of tangent is
\( y - y_1 = -\frac{x_1 - h}{y_1 - k}(x - x_1) \)
Substitute \(h = \frac{3}{2}\), \(k = -5\), \(x_1 = 4\), and \(y_1 = -11\) into the equation
\( y + 11 = -\frac{4 - \frac{3}{2}}{-11 + 5}(x - 4) \)
\( y + 11 = -\frac{\frac{5}{2}}{-6}(x - 4) \)
\( y + 11 = \frac{5}{12}(x - 4) \)
\( 5x - 12y = 152 \)
Next, the equation of normal is (perpendicular to the tangent)
\(12x+5y=k\)
The normal line passes through center \((h,k)=(\frac{3}{2}, -5)\), so the equation is
\(12x+5y+7=0\)
- \(x^2+y^2-8x-2y+12=0\) at (x, -1)
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Solution
Given equation of circle is
\(x^2 + y^2 - 3x + 10y - 15 = 0\)
To find the center and radius, we rewrite the equation in standard form, then
The center of the circle is
\((h, k) = (4,1)\)
The radius of the circle is
\(r^2 = 5 \)
Given point is
\((x_1, y_1) = (x,-1) \)
Now, the equation of tangent is
\( y - y_1 = -\frac{x_1 - h}{y_1 - k}(x - x_1) \)
\( y +1 = -\frac{x - 4}{-1 -1}(x - x) \)
\( y + 1 = 0 \)
Next, the equation of normal is (perpendicular to the tangent) is
\(x+k=0\)
The normal line passes through center \((h,k)=(4,1) \), so the equation is
\(x-4=0\)
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