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Condition for the line y = mx + c to be a tangent to the circle and finding the point of contact

**Condition for the line **** y
**=

Let the line *y *= *mx *+ *c *touch the circle *x*^{2}
+ *y*^{2} = *a*^{2} . The centre and radius of the
circle *x*^{2} + *y*^{2}
= *a*^{2} are (0, 0) and *a *respectively.

Then the perpendicular distance of the line *y *− *mx *−
*c *= 0 from (0, 0) is

This must be equal to radius .Therefore = *a *or *c*^{2}
= *a*^{2} (1+ *m*^{2}) .

**Thus the condition for the line **** y **=

Let (*x*_{1} , *y*_{1} ) be the the
point of contact of *y *= *mx *+ *c with the circle x*^{2}
+ *y*^{2} = *a*^{2},

Then *y*_{1} = *mx*_{1} + *c ……….*(1)

Equation of tangent at (*x*_{1} , *y*_{1}
) is *xx*_{1}* +
yy*_{1}* *= *a*^{2}
.

*yy*_{1}* *= -*xx*_{1}* *+ *a*^{2} ... (2)

Equations (1) and (2) represent the same line and hence the coefficients are proportional.

Then the points of contact is either

**Note**** **

The equation of tangent at P to a circle is *y *= *mx *±
*a √*[1* +m ^{2}*]

**Theorem 5.4**

From any point outside the circle *x*^{2} + *y*^{2}
= *a*^{2} two tangents can be drawn.

**Proof**

Let P(x_{1} , y_{1} ) be a point outside the
circle. The equation of the tangent is

It passes through (x_{1}, y_{1}) . Therefore

Squaring both sides, we get

(* y _{1} – mx_{1}*)

*y _{1}*

*m*^{2} (*x*_{1}^{2}
- *a*^{2} ) - 2*mx*_{1}*y*_{1}*
*+ ( *y*_{1}^{2}
- *a*^{2} ) = 0 .

This quadratic equation in *m *gives two values for *m *.

These values give two tangents to the circle *x*^{2}
+ *y*^{2} = *a*^{2}.

**Note**

(1) If (*x*_{1}, *y*_{1} ) is a point outside the circle, then both the
tangents are real.

(2) If (*x*_{1}, *y*_{1} ) is a
point inside the circle, then both the tangents are imaginary.

(3) If (*x*_{1}, *y*_{1} ) is a
point on the circle, then both the tangents coincide.

**Example 5.11**

Find the equations of the tangent and normal to the circle *x*^{2}
+ *y*^{2} = 25 at *P*(−3, 4) .

**Solution**

Equation of tangent to the circle at *P*(*x*_{1}
, *y*_{1} ) is *xx*_{1} + *yy*_{1} = *a*^{2}

That is, *x*(-3) + *y*(4) = 25

-3*x *+ 4 *y *= 25

Equation of normal is *xy*_{1} - *yx*_{1}
= 0

That is, 4*x *+ 3*y *= 0 .

**Example 5.12**

If *y *= 4*x *+ *c *is a tangent to the circle *x*^{2}
+ *y*^{2} = 9 , find *c*.

**Solution**

The condition for the line *y *= *mx *+ *c *to be
a tangent to the circle *x*^{2} + *y*^{2} = *a*^{2}
is *c*^{2} = *a*^{2} (1+ *m*^{2}).

Then, c = ± √[9(1+16)]

c = ±3 √17.

**Example 5.13**

A road bridge over an irrigation canal have two semi circular
vents each with a span of 20*m *and the supporting pillars of width 2*m *.
Use Fig.5.16 to write the equations that represent the semi-verticular vents.

**Solution**

Let O_{1} O_{2} be
the centres of the two semi circular vents.

First vent with centre *O*_{1} (12, 0) and radius *r *= 10 yields equation to first semicircle as

(*x *-12)^{2} + ( *y *- 0)^{2} = 10^{2}

Þ *x*^{2}
+ *y*^{2} - 24*x *+ 44 = 0 , *y *> 0 .

Second vent with centre *O*_{2} (34, 0) and radius *r
*= 10 yields equation to second vent as

(*x *- 34)^{2} + *y*^{2} = 10^{2}

Þ *x*^{2}
+ *y*^{2} - 68*x *+1056 = 0 , *y *> 0 .

Tags : Formula, Solved Example Problems , 12th Mathematics : UNIT 5 : Two Dimensional Analytical Geometry II

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12th Mathematics : UNIT 5 : Two Dimensional Analytical Geometry II : Condition for the line y = mx + c to be a tangent to the circle and finding the point of contact | Formula, Solved Example Problems

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