: If the tangent at $ (1,7) $ to the curve $ {x^2} = y - 6 $ touches the circle $ {x^2} + {y^2} + 16x + 12y + c = 0 $ then the value of $ c $ is
A. $ 85 $
B. $ 95 $
C. $ 195 $
D. $ 185 $
Answer
628.2k+ views
Hint:
Here we need to understand the given situation and the statement with the help of the diagram as it will make the situation clearer. First of all we will find the equation of the tangent to the given curve $ {x^2} = y - 6 $ as we need just the slope and the point which we have as $ (1,7) $
Once we get the equation of the tangent, we can say that the perpendicular distance from the centre of the circle to the tangent will be equal to the radius of the circle. Hence we can easily find the value of the unknown variable $ c $ .
Complete step by step solution:
Here we are given that the tangent at $ (1,7) $ to the curve $ {x^2} = y - 6 $ touches the circle $ {x^2} + {y^2} + 16x + 12y + c = 0 $ and we need to find the value of $ c $
So let us draw the figure for this problem.
As we know that tangent passes through the point $ (1,7) $ and it is tangent of the curve $ {x^2} = y - 6 $
So we can differentiate the above curve with respect to $ x $ and we can find the slope at the point $ (1,7) $
So differentiating the curve we get
$ \dfrac{d}{{dx}}{x^2} = \dfrac{d}{{dx}}y - \dfrac{d}{{dx}}6 $
$
2x = \dfrac{{dy}}{{dx}} - 0 \\
\dfrac{{dy}}{{dx}} = 2x \\
$
Hence we have got the slope which is $ \dfrac{{dy}}{{dx}} = 2x $
Now we can find the slope of this curve at the point $ (1,7) $ which will be
$ \dfrac{{dy}}{{dx}} = 2x $ $ = 2(1) = 2 $
Now we have a slope of the tangent as well as one passing point. So we can easily find the equation of the tangent which will be:
Point $ = (1,7) $
Slope $ = m = 2 $
So equation of tangent is written as
$
(y - 7) = m(x - 1) \\
y - 7 = 2(x - 1) \\
y - 7 = 2x - 2 \\
2x - y + 5 = 0 \\
$
Now we know that the circle whose general form is $ {x^2} + {y^2} + 2gx + 2fy + c = 0 $ has the point $ \left( { - g, - f} \right) $ as centre and the radius as $ \sqrt {({g^2} + {f^2} - c)} $
Hence we can compare this equation of the circle given with the general equation and get the value of the centre and radius which will be:
Equating the coefficient of $ x $ we get
$
2g = 16 \\
g = 8 \\
$
Equating the coefficient of $ y $ we get
$
2f = 12 \\
f = 6 \\
$
So we can say that the centre of the given circle is $ \left( { - g, - f} \right) = ( - 8, - 6) $
Radius $ = \sqrt {({g^2} + {f^2} - c)} = \sqrt {({8^2} + {6^2} - c)} = \sqrt {100 - c} $
Also we can say that the perpendicular distance of the centre from the point on the circumference is equal to the radius of the circle. Hence we can write that:
Distance from the centre to point $ (1,7) $ $ = \sqrt {100 - c} $
Perpendicular distance from the point $ (m,n) $ from the line $ ax + by + c = 0 $ is given by $ \dfrac{{am + bn + c}}{{\sqrt {{a^2} + {b^2}} }} $
Hence applying the same for the above we get:
Distance from the centre $ ( - 8, - 6) $ to tangent $ 2x - y + 5 = 0 $ $ = \sqrt {100 - c} $
$ \dfrac{{2( - 8) - ( - 6) + 5}}{{\sqrt {{2^2} + {1^2}} }} $ $ = \sqrt {100 - c} $
$ \left| {\dfrac{{ - 16 + 6 + 5}}{{\sqrt 5 }}} \right| = \left| {\dfrac{{ - 5}}{{\sqrt 5 }}} \right| = \left| { - \sqrt 5 } \right| = \sqrt 5 $ $ = \sqrt {100 - c} $
So we can say that $ \sqrt 5 $ $ = \sqrt {100 - c} $
Squaring both sides we get
$
5 = 100 - c \\
c = 100 - 5 = 95 \\
$
Hence B is the correct option.
Note:
In these types of questions we must know how we can calculate the radius and centre of the circle using the general equation of the circle. The circle whose general form is $ {x^2} + {y^2} + 2gx + 2fy + c = 0 $ has the point $ \left( { - g, - f} \right) $ as centre and the radius as $ \sqrt {({g^2} + {f^2} - c)} $
We must also know that perpendicular distance from the point $ (m,n) $ from the line $ ax + by + c = 0 $ is given by $ \dfrac{{am + bn + c}}{{\sqrt {{a^2} + {b^2}} }} $
Here we need to understand the given situation and the statement with the help of the diagram as it will make the situation clearer. First of all we will find the equation of the tangent to the given curve $ {x^2} = y - 6 $ as we need just the slope and the point which we have as $ (1,7) $
Once we get the equation of the tangent, we can say that the perpendicular distance from the centre of the circle to the tangent will be equal to the radius of the circle. Hence we can easily find the value of the unknown variable $ c $ .
Complete step by step solution:
Here we are given that the tangent at $ (1,7) $ to the curve $ {x^2} = y - 6 $ touches the circle $ {x^2} + {y^2} + 16x + 12y + c = 0 $ and we need to find the value of $ c $
So let us draw the figure for this problem.
As we know that tangent passes through the point $ (1,7) $ and it is tangent of the curve $ {x^2} = y - 6 $
So we can differentiate the above curve with respect to $ x $ and we can find the slope at the point $ (1,7) $
So differentiating the curve we get
$ \dfrac{d}{{dx}}{x^2} = \dfrac{d}{{dx}}y - \dfrac{d}{{dx}}6 $
$
2x = \dfrac{{dy}}{{dx}} - 0 \\
\dfrac{{dy}}{{dx}} = 2x \\
$
Hence we have got the slope which is $ \dfrac{{dy}}{{dx}} = 2x $
Now we can find the slope of this curve at the point $ (1,7) $ which will be
$ \dfrac{{dy}}{{dx}} = 2x $ $ = 2(1) = 2 $
Now we have a slope of the tangent as well as one passing point. So we can easily find the equation of the tangent which will be:
Point $ = (1,7) $
Slope $ = m = 2 $
So equation of tangent is written as
$
(y - 7) = m(x - 1) \\
y - 7 = 2(x - 1) \\
y - 7 = 2x - 2 \\
2x - y + 5 = 0 \\
$
Now we know that the circle whose general form is $ {x^2} + {y^2} + 2gx + 2fy + c = 0 $ has the point $ \left( { - g, - f} \right) $ as centre and the radius as $ \sqrt {({g^2} + {f^2} - c)} $
Hence we can compare this equation of the circle given with the general equation and get the value of the centre and radius which will be:
Equating the coefficient of $ x $ we get
$
2g = 16 \\
g = 8 \\
$
Equating the coefficient of $ y $ we get
$
2f = 12 \\
f = 6 \\
$
So we can say that the centre of the given circle is $ \left( { - g, - f} \right) = ( - 8, - 6) $
Radius $ = \sqrt {({g^2} + {f^2} - c)} = \sqrt {({8^2} + {6^2} - c)} = \sqrt {100 - c} $
Also we can say that the perpendicular distance of the centre from the point on the circumference is equal to the radius of the circle. Hence we can write that:
Distance from the centre to point $ (1,7) $ $ = \sqrt {100 - c} $
Perpendicular distance from the point $ (m,n) $ from the line $ ax + by + c = 0 $ is given by $ \dfrac{{am + bn + c}}{{\sqrt {{a^2} + {b^2}} }} $
Hence applying the same for the above we get:
Distance from the centre $ ( - 8, - 6) $ to tangent $ 2x - y + 5 = 0 $ $ = \sqrt {100 - c} $
$ \dfrac{{2( - 8) - ( - 6) + 5}}{{\sqrt {{2^2} + {1^2}} }} $ $ = \sqrt {100 - c} $
$ \left| {\dfrac{{ - 16 + 6 + 5}}{{\sqrt 5 }}} \right| = \left| {\dfrac{{ - 5}}{{\sqrt 5 }}} \right| = \left| { - \sqrt 5 } \right| = \sqrt 5 $ $ = \sqrt {100 - c} $
So we can say that $ \sqrt 5 $ $ = \sqrt {100 - c} $
Squaring both sides we get
$
5 = 100 - c \\
c = 100 - 5 = 95 \\
$
Hence B is the correct option.
Note:
In these types of questions we must know how we can calculate the radius and centre of the circle using the general equation of the circle. The circle whose general form is $ {x^2} + {y^2} + 2gx + 2fy + c = 0 $ has the point $ \left( { - g, - f} \right) $ as centre and the radius as $ \sqrt {({g^2} + {f^2} - c)} $
We must also know that perpendicular distance from the point $ (m,n) $ from the line $ ax + by + c = 0 $ is given by $ \dfrac{{am + bn + c}}{{\sqrt {{a^2} + {b^2}} }} $
Recently Updated Pages
Onehalf of a convex lens is covered with a black paper class 12 physics CBSE

Differentiate between lanthanoids and actinoids class 12 chemistry CBSE

An object 5 cm in length is held 25 cm away from a class 12 physics CBSE

Name the following halides according to the IUPAC system class 12 chemistry CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

An infinite ladder network of resistances is constructed class 12 physics CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples of unisexual and bisexual flowers

